Projection of Vector on a line or direction MCQ Quiz in मराठी - Objective Question with Answer for Projection of Vector on a line or direction - मोफत PDF डाउनलोड करा

Explanation:

The additive inverse vector of vector pi + 2j - 3k is

-pi - 2j + 3k

So, $(\sqrt{\mathrm{a}\mathrm{b}}\mathrm{i}+\sqrt{\mathrm{b}}\mathrm{j}+\sqrt{\mathrm{p}}\mathrm{k})$ = -pi - 2j + 3k

Comparing both sides

$\sqrt{ab}=-p,\sqrt{b}=-2,\sqrt{p}=3$

So, p = 9, b = 4

Putting in $\sqrt{ab}=-p$ we get

$\sqrt{4b}=-9$

i.e., 4b = 91

i.e., b = 81/4

Hence p = 9, b = 4, a = 81/4

Option (2) is true.

$\therefore$ Required magnitude of projection ${\displaystyle \frac{|(2\hat{i}+3\hat{j}+\hat{k}).(\hat{i}-2\hat{j}+\hat{k})|}{|\hat{i}-2\hat{j}+\hat{k}|}}$

${\displaystyle \frac{|2-6+1|}{\sqrt{6}}}={\displaystyle \frac{3}{\sqrt{6}}}=\sqrt{{\displaystyle \frac{3}{2}}}$

Concept Used:

Projection of vector $a$ on $b$ is $\frac{(a.b)}{|b|}$

Calculation

Given:

$A(0,3,-3)$, $B(1,1,1)$, $C(2,0,3)$

$AB=B-A=(1-0,1-3,1-(-3))=(1,-2,4)$

$AC=C-A=(2-0,0-3,3-(-3))=(2,-3,6)$

$AB.AC=(1\times 2)+(-2\times -3)+(4\times 6)=2+6+24=32$

$|AC|=\sqrt{2^2+(-3{)}^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$

Projection of AB on AC = $\frac{(AB.AC)}{|AC|}=\frac{32}{7}$

$\therefore$ Projection of AB on AC = $\frac{32}{7}$

Hence option 2 is correct

Answer : 3

Solution :

$\overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

$\overline{\mathrm{CD}}=3\hat{\mathrm{i}}-6\hat{\mathrm{j}}+2\hat{\mathbf{k}}$

Vector projection of $\overline{\mathrm{AB}}$ on $\overline{\mathrm{CD}}$ 

= $(\overline{\mathrm{AB}}\cdot \overline{\mathrm{CD}})\frac{\overline{\mathrm{CD}}}{|\overline{\mathrm{CD}}{|}^2}$ 

= $(-3+6-4)\frac{(3\hat{\mathrm{i}}-6\hat{\mathrm{j}}+2\hat{\mathrm{k}})}{{\left(\sqrt{3^2+(-6{)}^2+2^2}\right)}^2}$ 

= $\frac{-1}{49}(3\hat{i}-6\hat{j}+2\hat{k})$

Concept:

Let $\overrightarrow{u}and\overrightarrow{v}$ be two vectors. Then the vector component of the vector $\overrightarrow{u}on\overrightarrow{v}$ is given by:

 $\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\left|\overrightarrow{v}\right|}(\hat{v})=\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{{\left|\overrightarrow{v}\right|}^2}(\overrightarrow{v})$

Calculation:

Given:

$\overrightarrow{a}=4\hat{i}+6\hat{j}$ and $\overrightarrow{b}=3\hat{j}+4\hat{k}$

$\overrightarrow{a}.\overrightarrow{b}=(4,6,0).(0,3,4)$

$\overrightarrow{a}.\overrightarrow{b}=18$

|$\overrightarrow{a}$| = √52 

$|\overrightarrow{b}|=5$ 

The component of vector a along b is $\frac{18}{25}(3j+4k)$

Mistake PointsWe are asked to find the vector component of 'a' along 'b'. If the scalar component was asked, then there would have been only 5 in the denominator.

Explanation:

The projection vector of $\overrightarrow{\mathrm{a}}$ on $\overrightarrow{\mathrm{b}}$ is given by $\frac{\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}$.

The correct answer is option 2.

Concept:

The projection of vector $\overrightarrow{\mathrm{a}}$  on $\overrightarrow{\mathrm{b}}$ is given by $\frac{\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{b}|}$.

Calculation:

Given  $\overrightarrow{\mathrm{a}}$ = 2î − ĵ + k̂  and  $\overrightarrow{\mathrm{b}}$ = î + 2ĵ + 2k̂

 $\overrightarrow{a}\cdot \overrightarrow{b}$ = ( 2î − ĵ + k̂ )⋅ ( î + 2ĵ + 2k̂ )

= 2 × 1 + (-1) × 2 + 1 × 2 

= 2 - 2 + 2

= 2

|$\overrightarrow{\mathrm{b}}$| = | î + 2ĵ + 2k̂ |

= $\sqrt{1^2+2^2+2^2}$

= $\sqrt{1+4+4}=\sqrt{9}$ 

= 3

The projection of vector $\overrightarrow{\mathrm{a}}$  on $\overrightarrow{\mathrm{b}}$  = $\frac{\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{b}|}$.

= $\frac{2}{3}$

The projection of vector $\overrightarrow{\mathrm{a}}$ = 2î − ĵ + k̂ along $\overrightarrow{\mathrm{b}}$ = î + 2ĵ + 2k̂ is 2/3.

The correct answer is option 1.

Concept -

Use of normal vector  and projection vector

projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is $=|\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|}|$

Solution - 

Normal vector to the plane to plane containing $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$ is 

$\overrightarrow{n}=(\hat{i}+\hat{j}+\hat{k})\times (\hat{i}+2\hat{j}+3\hat{k})$

 $\overrightarrow{n}=(\hat{i}-2\hat{j}+\hat{k})$

Projection of $(2\hat{i}+3\hat{j}+\hat{k})$ on $\overrightarrow{n}$ = $=|\frac{(2\hat{i}+3\hat{j}+\hat{k}).(\hat{i}-2\hat{j}+\hat{k})}{\sqrt{1+4+1}}|$

= $\frac{3}{\sqrt{6}}=\sqrt{\frac{3}{2}}$

Hence the final answer is option 2.