Three Dimensional Geometry Class 12 NCERT Exemplar Solutions -...

Concept used. Direction cosines parametrise the unit sphere of directions. Any vector of length r along that direction is r times the corresponding unit vector.

1. Identify the two known direction cosines from the given angles: cos 60^∘=12 and cos 45^∘=1√2.
2. Apply the identity l²+m²+n²=1 to find the magnitude of the missing DC.
3. Both signs are admissible because the constraint involves squares; geometrically the line has two senses, one going above the xy-plane and one going below.
4. Multiply the resulting unit vector by the prescribed length to obtain the position vector.

Common alternative. Some students first convert the magnitudes (cos 60^∘,cos 45^∘) into Cartesian components (5, 5√2) for the x- and y-components directly, then determine the z-component from √|OA⃗|²-x²-y²=√100-25-50=5. This avoids using l,m,n explicitly and is acceptable in CBSE marking schemes.

Why the ±. The angles with two axes do not pin down a single direction in 3D; the third coordinate can be positive or negative. Always present both.

OA⃗=5î+5√2 ĵ± 5k̂.

Concept used. Two pieces of information determine a unique line in space: one point on the line, plus a direction vector. The vector form is the cleanest way to encode both.

1. Write the position vector of the given point as a⃗.
2. Use the given parallel vector as the direction vector b⃗.
3. Assemble r⃗=a⃗+λb⃗ with λ a real parameter.

Cartesian form sanity check. Reading off the components of b⃗=(3,-2,6) as direction ratios, the equivalent symmetric Cartesian form is x-13=y+2-2=z-36, which is what CBSE often asks for in the same question.

Marking-scheme angle. Examiners look for (i) the standard-form identification of a⃗ and b⃗, (ii) the assembled vector equation, and (iii) a clean line of working with λ as the scalar parameter. Two marks for this question is typical.

r⃗=(î-2ĵ+3k̂)+λ(3î-2ĵ+6k̂).

Concept used. A pair of lines in 3D has three possible relationships - intersecting, parallel, or skew. Intersection is verified by exhibiting a common point.

1. Convert both symmetric forms to parametric coordinates, using different scalar parameters (λ for one, μ for the other).
2. Set up the three coordinate-matching equations.
3. Two equations suffice to solve for (λ,μ); the third equation is a consistency check.
4. If consistent, substitute back to obtain the intersection point.

Coplanarity test (alternative). The two lines are coplanar (intersecting or parallel) iff (b₁⃗×b₂⃗)·(a₂⃗-a₁⃗)=0. Here b₁⃗=(2,3,4), b₂⃗=(5,2,1), and a₂⃗-a₁⃗=(3,-1,-3). Compute b₁⃗×b₂⃗=(3·1-4·2, 4·5-2·1, 2·2-3·5)=(-5,18,-11). Dot with (3,-1,-3): -15-18+33=0. So coplanar, and since direction vectors are not proportional, the lines intersect.

Marking-scheme angle. Show the parametric set-up, solve cleanly, verify in the third equation, and state the intersection point. Skipping verification often costs 1 mark.

The two lines intersect at (-1,-1,-1).

Concept used. The angle between two lines is the angle between their direction vectors (up to the acute-angle convention).

1. Read direction vectors b₁⃗ and b₂⃗ off the vector forms.
2. Compute the dot product. Take the absolute value to ensure the acute angle.
3. Compute the two magnitudes.
4. Divide and take cos^-1.

Numerical sanity check. 1921≈ 0.905, so θ≈ 25.2^∘. This is a small acute angle, consistent with two direction vectors that share two positive component signs and have similar magnitudes.

Common slip. Dropping the modulus on the dot product gives the obtuse supplement when the dot product is negative. CBSE wants the acute angle, so always include the modulus.

θ=cos^-1(1921).

Concept used. Coplanarity of two lines is equivalent to a vanishing scalar triple product of (direction-1, direction-2, joining vector). When two lines are coplanar but not parallel, they meet in exactly one point.

1. Express the two lines through pairs of points A,B and C,D.
2. Form direction vectors AB⃗ and CD⃗, and a joining vector AC⃗.
3. Compute the scalar triple product as a determinant.
4. Show it is zero (coplanar) and that the direction vectors are not proportional (not parallel). Conclude intersection.

Finding the point of intersection (optional). Parametrise line AB as (0,-1,-1)+λ(4,6,2) and line CD as (3,9,4)+μ(-7,-5,0). Equating gives 4λ+7μ=3, 6λ+5μ=10, 2λ=5. From the third, λ=52; substituting in the first gives 10+7μ=3, so μ=-1. Check in the second: 6(52)+5(-1)=15-5=10. Intersection point is (0,-1,-1)+52(4,6,2)=(10,14,4).

Marking-scheme angle. Examiners award separate credit for the triple-product set-up, the determinant evaluation, the parallel-vs-non-parallel observation, and the final conclusion sentence.

The lines are coplanar (triple product =0) and non-parallel, hence they intersect.

Concept used. When a line is given as the intersection of two planes (here x-py-q=0 and z-ry-s=0), the simplest way to read off its direction is to parametrise by the variable that is free.

1. Choose y as the parameter because both equations are linear in y.
2. Read off the direction-vector components as ∂(x,y,z)/∂ y.
3. Set the dot product equal to zero to get the perpendicularity condition.

Geometric note. The middle "1" in the dot product comes from the ∂ y/∂ y = 1 component, which both lines share. The condition pp'+rr'+1=0 therefore says that the two perpendicular contributions (in x and z) cancel the unavoidable "1" from the shared y-direction.

Cross-product alternative. Each line can be written as the intersection of planes; its direction vector equals the cross product of the two plane normals. For line 1, normals are (1,-p,0) and (0,-r,1), cross product (-p,-1,-r) - same direction up to sign as (p,1,r).

pp'+rr'+1=0 is the required perpendicularity condition.

Concept used. The set of points equidistant from A and B is the perpendicular bisector plane of the segment AB.

1. Locate the mid-point M of AB, which lies on the required plane.
2. Take AB⃗ as the normal vector to the plane (the plane is perpendicular to the segment).
3. Use the point-normal form n⃗·(r⃗-M⃗)=0.
4. Simplify to obtain the Cartesian equation.

Equidistance verification. For any (x,y,z) on x+y+2z=19, the squared distance to A(2,3,4) minus the squared distance to B(4,5,8) expands to a linear expression that vanishes exactly on the plane. Plugging in M=(3,4,6) gives |MA|²=1+1+4=6=|MB|², confirming equidistance.

Marking-scheme angle. CBSE awards credit for (i) the mid-point, (ii) the normal vector, (iii) the point-normal equation, and (iv) the simplified Cartesian form. Each is one mark for a 4-mark question.

x+y+2z=19.

Concept used. The normal form n̂·r⃗=p encodes the perpendicular distance from the origin to the plane as p, provided n̂ is a unit vector.

1. "Equally inclined to the axes" forces all three direction cosines equal.
2. The unit-vector constraint pins each to ±1√3.
3. Set up n̂·r⃗=p and clear the surd.

Geometric sanity check. The point (3,3,3) lies on the plane x+y+z=9 and its distance from the origin is √27=3√3. Since the normal (1,1,1)/√3 points from the origin to (3,3,3), this matches the prescribed perpendicular distance exactly.

x+y+z=± 9.

Concept used. A normal vector + a point on the plane uniquely determines the plane.

1. Form the direction vector along the perpendicular line.
2. Take this vector as the plane's normal.
3. Use the foot of the perpendicular as the on-plane point.
4. Plug into the point-normal form and simplify.

Distance verification. Distance from (-2,-1,-3) to 3x-2y+6z-27=0 is |3(-2)-2(-1)+6(-3)-27|√9+4+36=|-6+2-18-27|7=497=7. Distance from (-2,-1,-3) to (1,-3,3) is √9+4+36=7. The two match, confirming the foot of perpendicular.

3x-2y+6z=27.

Concept used. Three non-collinear points determine a unique plane. Two edge vectors from a common vertex span the plane; their cross product is normal to the plane.

1. Pick one vertex, say P, as the on-plane point.
2. Compute PQ⃗ and PR⃗ - two non-parallel in-plane vectors.
3. Cross-product them to get a normal.
4. Use the point-normal form and simplify.

Determinant alternative. The Cartesian plane through (x_i,y_i,z_i), i=1,2,3, can also be written as vmatrix x-x₁ & y-y₁ & z-z₁ x₂-x₁ & y₂-y₁ & z₂-z₁ x₃-x₁ & y₃-y₁ & z₃-z₁vmatrix=0, which expands to the same Cartesian equation. Use whichever you find faster.

7x+3y-z=17.

Concept used. A one-parameter family of lines from the origin, each intersecting a given line, is the standard set-up; the angle condition selects discrete members from the family.

1. Parametrise points on the given line by t.
2. Treat (3+2t, 3+t, t) as the direction ratios of the required line through the origin.
3. Set cos(π/3)=12 equal to the cosine of the angle formula.
4. Solve the resulting quadratic in t. Each root gives one of the two required lines.

Algebraic simplification. The shortcut here is to notice S=6(t²+3t+3), which makes √6√S=6√t²+3t+3 - a clean factorisation. Spotting it early avoids cubic-looking algebra and keeps the final quadratic linear in disguise.

x1=y2=z-1 and x-1=y1=z-2.

Concept used. Two equations in three direction cosines (one linear, one quadratic) generically give two discrete lines. The angle between them is then a direct cosine computation.

1. Use the linear equation to eliminate one DC.
2. Substitute into the quadratic to get a ratio condition on the remaining two DCs.
3. Solve to obtain the two direction-ratio triples.
4. Compute the angle using the standard cosine formula.

Symmetry note. The two solutions (0,1,-1) and (1,0,-1) are related by swapping the roles of l and m, which is consistent with the symmetry of the original system in l and m.

θ=π3.

Concept used. Differentiating the unit-norm constraint û·û=1 gives û· dû=0, i.e. infinitesimal changes in a unit vector are perpendicular to the vector itself.

1. Identify both endpoints as unit vectors.
2. Expand |û₂|²=1 to leading order in the small quantities δ l,δ m,δ n.
3. Compute the dot product to second order.
4. Compare with the small-angle expansion of cosθ.

Geometric picture. The two adjacent directions û₁ and û₂ are nearby points on the unit sphere; the small displacement vector δu⃗=(δ l,δ m,δ n) is approximately tangent to the sphere, and its squared length |δu⃗|² equals the squared arc length θ² to leading order.

θ²=δ l²+δ m²+δ n².

Concept used. The DCs of any line through the origin are obtained by normalising its endpoint vector. A plane normal to that line through any specified point is given by the point-normal form.

1. Compute |OA|.
2. Divide (a,b,c) component-wise by |OA| to get the DCs.
3. Use (a,b,c) as the plane's normal and (a,b,c) as the on-plane point in the point-normal form.
4. Simplify to obtain the Cartesian equation.

Distance from origin to plane. The plane ax+by+cz=a²+b²+c² has perpendicular distance from the origin equal to |a²+b²+c²|√a²+b²+c²=√a²+b²+c²=|OA|. Consistent with the plane being normal to OA at the point A.

DCs =(a,b,c)√a²+b²+c², plane: ax+by+cz=a²+b²+c².

Concept used. A rotation of axes preserves Euclidean distance, so any rotation-invariant quantity expressed in coordinates of one system equals the same expression in coordinates of the rotated system.

1. Use the intercept-form perpendicular-distance formula for both coordinate systems.
2. Note that the geometric distance from the origin to the plane is the same in both systems.
3. Equate the two expressions for 1/p².

Why the intercept-distance formula works. The plane xa+yb+zc=1 has normal (1/a,1/b,1/c) and constant term 1, so the distance from the origin is |1|√1/a²+1/b²+1/c². Inverting and squaring gives 1p²=1a²+1b²+1c².

1a²+1b²+1c²=1a'²+1b'²+1c'² (invariance of origin-to-plane distance).

Concept used. The foot of perpendicular from an external point to a line is the point on the line closest to the external point. The connecting vector is perpendicular to the line's direction vector.

1. Bring the line equation to standard symmetric form.
2. Parametrise points on the line.
3. Impose perpendicularity of the connecting vector to the direction vector.
4. Solve the resulting linear equation for t.
5. Substitute back to get the foot and compute the distance.

Cross-product distance check. The distance from P to the line through A=(4,0,1) with direction b⃗=(-2,6,-3) can also be computed as |AP⃗×b⃗||b⃗|. Here AP⃗=(-2,3,-9) and AP⃗×b⃗=(-9· 6-(-3)· 3, (-9)(-2)-(-2)(-3), (-2)(6)-3(-2))=(-45,12,-6). Wait, let me recompute: AP⃗=(2-4, 3-0, -8-1)=(-2,3,-9). AP⃗×b⃗=(3(-3)-(-9)(6), (-9)(-2)-(-2)(-3), (-2)(6)-3(-2))=(-9+54, 18-6, -12+6)=(45,12,-6). Magnitude =√2025+144+36=√2205=21√5. Divide by |b⃗|=√4+36+9=7. Distance =21√57=3√5.

Foot: (2,6,-2), distance =3√5.

Concept used. The cross-product distance formula is the cleanest way to compute the perpendicular distance from a point to a line. No parametrisation is needed.

1. Read off a point on the line (A) and the direction vector (b⃗) from the symmetric form.
2. Form the vector AP⃗ from A to the external point.
3. Compute AP⃗×b⃗ via the 3× 3 determinant.
4. Divide the magnitude by |b⃗|.

Arithmetic tip. When the cross product comes out as (-35,56,21), notice the common factor 7 before squaring. Factoring early keeps the numbers small.

d=7.

Concept used. Standard plane-distance and foot-of-perpendicular formulas are direct substitutions; the main task is bookkeeping.

1. Identify A,B,C,D in the plane equation.
2. Plug the external point into Ax₀+By₀+Cz₀+D.
3. Divide by √A²+B²+C² for the distance.
4. Use F=P-(scale) (A,B,C) for the foot, where scale = (numerator)/(denominator-squared).

Verification. Always plug the computed foot back into the plane equation. Any non-zero residual signals an arithmetic slip earlier in the calculation.

Distance √6, foot (0,5/2,0).

Concept used. A line parallel to a plane has direction perpendicular to that plane's normal. Parallel to two planes therefore means direction perpendicular to both normals - exactly the cross product.

1. Read off the normal of each plane.
2. Cross-product them to obtain the line's direction.
3. Use the given point and the new direction in the line equation.

Sanity check. Direction (-2,1,3) dotted with n₁⃗=(1,2,0) gives -2+2+0=0. Dotted with n₂⃗=(0,3,-1) gives 0+3-3=0. Confirms perpendicularity to both normals.

r⃗=3î+k̂+λ(-2î+ĵ+3k̂).

Concept used. Two planes are perpendicular iff one's normal lies in the other plane. So the normal of the perpendicular plane is an in-plane direction of the required plane.

1. Form the in-plane vector from the two given points.
2. Use the normal of the perpendicular plane as a second in-plane vector.
3. Cross-product them to get the required plane's normal.
4. Plug into the point-normal form.

Verification. The required plane's normal (18,17,4) dotted with the perpendicular plane's normal (1,-2,4): 18-34+16=0. The two normals are perpendicular, confirming the two planes are perpendicular.

18x+17y+4z=49.

Concept used. The cross-product distance formula for skew lines is the high-yield 5-mark template in Chapter 11. Drilling it on a numerically clean example like this is the best preparation for the board exam.

1. Identify the two position vectors and the two direction vectors.
2. Compute the difference of position vectors.
3. Compute the cross product of direction vectors.
4. Take the absolute value of the dot product of (cross product) with (difference of positions).
5. Divide by the magnitude of the cross product.

Skew vs parallel check. b₁⃗×b₂⃗=12(2,3,6)≠0⃗, so the two lines are not parallel. The non-zero numerator confirms they do not intersect either. The lines are skew, and the shortest distance is finite and positive.

d=14.

Concept used. The family P₁+λ P₂=0 generates every plane through the line of intersection of P₁=0 and P₂=0. One scalar condition (here perpendicularity) pins down a unique member.

1. Write the parameter family.
2. Read off the normal of the parametric plane.
3. Impose the perpendicularity condition with the given plane.
4. Solve for λ and substitute back.

Verification. The normal (51,15,-50) dotted with (5,3,6): 255+45-300=0. Confirms perpendicularity to the prescribed plane.

51x+15y-50z+173=0.

Concept used. A rotation about the line of intersection is equivalent to varying the parameter λ in the family P₁+λ P₂=0. The angle between the rotated and the original plane is a function of λ.

1. Write the rotated plane as ax+by+λ z=0.
2. Compute the angle between the rotated and original plane via the cosine formula.
3. Set this cosine equal to cosα and solve for λ.
4. Both signs of λ are valid because rotation can be clockwise or counter-clockwise.

Sanity check at α=0. tan 0=0, so λ=0 and the equation reduces to ax+by=0 - the original plane, as expected.

Sanity check at α=π/2. tan(π/2) diverges, meaning the plane perpendicular to the original (through the same line of intersection) is z=0 - which is exactly P₂. Consistent.

ax+by±√a²+b² tanα · z=0.

Concept used. The unit-distance condition pins λ as a real number through a quadratic, generically giving two valid planes.

1. Write the family P₁+λ P₂=0 in Cartesian form.
2. Distance from the origin is |D|/√A²+B²+C² where A,B,C,D are the coefficients.
3. Set equal to 1, square, and solve for λ.
4. Substitute each λ back to get the two planes.

Verification. For 2x+y-2z=3, distance from origin =|3|√4+1+4=33=1. For x-2y-2z=-3, distance =|-3|√1+4+4=33=1.

Two planes: 2x+y-2z=3 or x-2y-2z=-3.

Concept used. The signed distance encodes both magnitude and side. Opposite signs of the signed distance correspond to opposite sides of the plane.

1. Plug each point into the LHS of the plane equation (no modulus yet).
2. Compare absolute values for "equidistant".
3. Compare signs for "opposite sides".

Mid-point lies on the plane. The mid-point of P₁P₂ is (2,1,3). Plug in: 5(2)+2(1)-7(3)+9=10+2-21+9=0. So the mid-point lies on the plane - exactly what equidistant points on opposite sides give.

Equidistant: 9√78; opposite sides confirmed by sign reversal of the signed distance.

Concept used. The common perpendicular of two skew (or coplanar) lines is the segment connecting the closest pair of points on the two lines. Its direction is parallel to b₁⃗×b₂⃗.

1. Parametrise P on line 1 and Q on line 2.
2. Write PQ⃗ in terms of the two parameters.
3. Set PQ⃗·b₁⃗=0 and PQ⃗·b₂⃗=0.
4. Solve the resulting 2× 2 linear system for (λ,μ).
5. Substitute back to find P and Q.

Cross-product cross-check. AB⃗×CD⃗=(3,-1,1)×(-3,2,4)=((-1)(4)-1(2), 1(-3)-3(4), 3(2)-(-1)(-3))=(-6,-15,3)=3(-2,-5,1). The vector PQ⃗=Q-P=(-3-3, -7-8, 6-3)=(-6,-15,3)=3(-2,-5,1). PQ⃗ is indeed parallel to AB⃗×CD⃗, confirming it is the common perpendicular.

P=(3,8,3), Q=(-3,-7,6).

Concept used. A linear + quadratic system in three DCs generically gives two discrete lines. Their relative angle (perpendicularity in particular) is read off from the dot product of their direction ratios.

1. Use the linear equation to eliminate one DC.
2. Substitute and factor the quadratic in the remaining two.
3. Read off the two solution direction-ratio triples.
4. Check perpendicularity by summing products of DRs.

Symmetry insight. The two solutions (1,-2,-2) and (-2,1,-2) are related by swapping the roles of l and m - consistent with the symmetry of 2l+2m-n=0 and mn+nl+lm=0 under l↔ m.

The lines are perpendicular (dot product of DRs = 0).

Concept used. An orthonormal frame in 3D is a rotation of the standard î,ĵ,k̂ frame, and the sum e₁̂+e₂̂+e₃̂ in any orthonormal frame is a diagonal-type vector that makes equal angles with each basis vector.

1. Use the orthonormal-frame property: pairwise dot products vanish, individual squared norms are 1.
2. Compute |u⃗|² by expanding the sum.
3. Compute u⃗·e_î - only one term survives orthogonality.
4. Conclude the cosine of the angle is 1/√3 for every i.

Geometric picture. The vector î+ĵ+k̂ is the body diagonal of a unit cube and makes equal angles cos^-1(1/√3)≈ 54.7^∘ with all three axes. The result here is the same statement in an arbitrary orthonormal frame, by rotational invariance.

Equal angles cos^-1(1/√3) with each of the three perpendicular lines.

Concept used. Distance to a coordinate axis is found by dropping the coordinate along that axis and taking the magnitude of what remains.

1. Drop β (the y-coordinate).
2. Magnitude of (α,0,γ) is √α²+γ².

General rule. Distance from (α,β,γ) to x-axis is √β²+γ²; to y-axis is √α²+γ²; to z-axis is √α²+β².

(D).

Concept used. The DC identity l²+m²+n²=1 is the only constraint - it allows k to be either positive or negative.

1. Apply the identity with three equal DCs.
2. Solve the simple square-root equation.
3. Include both signs because a line has two senses.

(D).

Concept used. When the plane is given in the normal form r⃗·n̂=p with n̂ a unit vector, p is directly the distance from the origin.

1. Check whether n⃗ is already a unit vector.
2. If yes, distance is just |d|.
3. Here |n⃗|=1 exactly, so distance is 1.

(A).

Concept used. Mnemonic: line-vs-plane angle uses sin; line-vs-line angle uses cos. The reason is that the line-plane angle is complementary to the line-normal angle.

1. Identify b⃗ and n⃗.
2. Compute the dot product and magnitudes.
3. Use sinθ=|b⃗·n⃗|/(|b⃗||n⃗|).

Rationalisation. 15√2=15√2·√2√2=√210 - which is option (D).

(D).

Concept used. Reflection in a coordinate plane negates only the coordinate perpendicular to that plane.

1. xy-plane is z=0, so reflection flips the z-coordinate.

(D).

Concept used. Cross-product magnitude gives the area of a parallelogram spanned by two adjacent edge vectors.

1. Verify ABCD is a parallelogram by checking AB⃗=DC⃗.
2. Compute two adjacent edge vectors from a common vertex.
3. Cross-product and take magnitude.

(A).

Concept used. A homogeneous equation that factors into two linear factors represents a union of two planes through the origin.

1. Factor the second-degree expression.
2. Read off the two planes from the factors.
3. Check perpendicularity by dotting the two normals.

(D).

Concept used. The x-axis is the line through the origin with direction î=(1,0,0).

1. Plug b⃗=(1,0,0) into the line-plane angle formula.
2. Compute and simplify.

(C).

Concept used. Intercept form is the fastest plane equation when all three axis-intercepts are given.

1. Read off the three intercepts.
2. Plug into the intercept formula.
3. Multiply through by the LCM of denominators for a clean integer form.

6x+4y+3z=12.

Concept used. Direction cosines = direction ratios divided by the vector's magnitude.

1. Compute the magnitude.
2. Divide each component by it.

(23,23,-13).

Concept used. Symmetric Cartesian and vector forms are equivalent; converting is mechanical.

1. Read the point from the numerators (sign-flipped).
2. Read the direction vector from the denominators.
3. Assemble.

r⃗=(5î-4ĵ+6k̂)+λ(3î+7ĵ+2k̂).

Concept used. Two-point vector form. Either point may serve as a⃗; the direction b⃗-a⃗ comes out with the sign convention chosen.

1. Pick one point as the base a⃗.
2. Use the difference vector as the direction.

r⃗=(3î+4ĵ-7k̂)+λ(-2î-5ĵ+13k̂).

Concept used. Vector-form to Cartesian conversion is a direct dot product.

1. Expand the dot product component-wise.
2. Equate to the RHS scalar.

x+y-z=2.

Concept used. The plane's normal vector is read off directly from the coefficients of x,y,z in the plane equation. Unit normal = normal / its magnitude.

1. Read coefficients.
2. Normalise.

True.

Concept used. Intercept form xa+yb+zc=1 exposes the intercepts directly.

1. Move the constant to the RHS.
2. Divide every term by the new RHS.
3. Read off a,b,c as the reciprocals of the coefficients.

True.

Concept used. Line-plane angle uses sin on the dot product of (line direction) and (plane normal), normalised by the two magnitudes.

1. Recompute carefully.
2. Compare to the stated answer.

Numerical recheck. b⃗·n⃗=2(3)+(-1)(-4)+(1)(-1)=6+4-1=9. |b⃗||n⃗|=√6· 26=√156=2√39. sinθ=92√39. The stated value 52√91 differs - so the claim is False.

False.

Concept used. Plane-plane angle uses cos on the dot product of normals.

1. Read the two normals.
2. Apply the cosine-of-angle formula.

True.

Concept used. Two conditions are needed for a line to lie in a plane.

1. Check whether a point of the line lies on the plane.
2. If not, the line cannot lie in the plane regardless of direction.

Common slip. Students sometimes only check perpendicularity of direction vs normal and conclude "the line lies in the plane". That condition only makes the line parallel to the plane (or in it); the on-plane point check is what distinguishes the two cases.

False.

Concept used. Mechanical conversion between Cartesian symmetric and vector parametric forms.

1. Read the point.
2. Read the direction.
3. Compare.

True.

Concept used. DR sign mistakes are the most common error in writing line equations. Always copy the parallel-vector components verbatim.

1. Re-derive the equation.
2. Spot the discrepancy in the y-denominator sign.

False.

Concept used. The vector from origin to the foot of perpendicular is normal to the plane and its length equals the perpendicular distance from the origin to the plane.

1. Identify OF⃗ as the plane's normal.
2. Use r⃗·OF⃗=|OF⃗|² (because F itself lies on the plane and F⃗·F⃗=|F⃗|²).

True.