Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

Match List-I with List-II

List-I

List-II

 A. 

 Distance of the point (1, -1,  1) from the plane $\vec{r} . (\hat{i}-\hat{j}+\hat{k})=2$, is 

 I. 

$x+y+z=2$

 B. 

 $\vec{r} . (\hat{i}+\hat{j}+\hat{k})=2$

 II. 

$-x+y+z=5$

 C. 

 Distance of a point (1, -1,  1) from the plane $\vec{r} . (\hat{i}+\hat{j}-\hat{k})=5$, is

 III. 

$\frac{1}{\sqrt{3}}$

 D. 

 $\vec{r} . (-\hat{i}+\hat{j}+\hat{k})=5$,

 IV. 

$2 \sqrt{3}$

Choose the correct answer from the options given below.

Options:

A-IV, B-II, C-III, D-I

A-III, B-II, C-IV, D-I

A-IV, B-I, C-III, D-II

A-III, B-I, C-IV, D-II

Correct Answer:

A-III, B-I, C-IV, D-II

Explanation:

A. $d=\left|\frac{1+1+1-2}{\sqrt{1^2+(-1)^2+1^2}}\right|=\frac{1}{\sqrt{3}}$        (III)

B. $\vec{r} . (\hat{i}+\hat{j}+\hat{k})=2 \Rightarrow x+y+z=2$           (I)

C. $d=\left|\frac{1-1-1-5}{\sqrt{1^2+1^2+(-1)^2}}\right|=\frac{6}{\sqrt{3}}=2 \sqrt{3}$        (IV)

D. $\vec{r} . (-\hat{i}+\hat{j}+\hat{k})=5 \Rightarrow -x+y+z=5$        (II)

Option: D