NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2

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Class 11 Maths NCERT Solutions Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2 is based on Distance between Two Points.

Download PDF NCERT Solutions for Class 12 Introduction to Three Dimensional Geometry Exercise 12.2

Check out the solutions of Class 11 Maths NCERT Solutions Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

Read More: NCERT Solutions For Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Also check other Exercise Solutions of Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

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Chapter Related Topics
Angle Between Two Plane	Angle between two lines	Angle between line and plane

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CBSE CLASS XII Related Questions

1.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]
- Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
2.
If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]
View Solution
3.
For a square matrix \(A\), \[ (3A)^{-1}= \]
- \( 3A^{-1} \)
- \( 9A^{-1} \)
- \( \frac{1}{3} A^{-1} \)
- \( \frac{1}{9} A^{-1} \)
View Solution
4.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:
- \( e^{-3} \)
- \( -1 \)
- \( 1 \)
- \( -e^3 \)
View Solution
5.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:
- \(\frac{1}{3}\)
- \(\frac{1}{9}\)
- \(3\)
- \(9\)
View Solution
6.
A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.
View Solution

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