Tag: applications of vector algebra

Questions Related to applications of vector algebra

If the line $\cfrac{x-1}{2}=\cfrac{y+3}{1}=\cfrac{z-5}{-1}$ is parallel to the plane $px+3y-z+5=0$, then the value of $p$

  1. $2$

  2. $-2$

  3. $\cfrac{1}{2}$

  4. $\cfrac{1}{3}$


Correct Option: B
Explanation:
line $11$ plane 
$\therefore$ line $\bot$ normal to plane 
$\therefore (2)(P)+(1)(3)+(-1)(-1)=0$  
$\therefore 2p + 3 + 1 =0$
$\therefore P=-2$

The angle between the plane $2 x - y + z = 6$ and a perpendiculars to the planes $x + y + 2 z = 7$ and $x - y = 3$ is

  1. $\frac { \pi } { 4 }$

  2. $\frac { \pi } { 3 }$

  3. $\frac { \pi } { 6 }$

  4. $\frac { \pi } { 2 }$


Correct Option: D

Statement 1: Line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$.
Statement 2: If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.


Correct Option: A
Explanation:

If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$ &  $\vec a\cdot \vec c=n$
Therefore, statement 2 is true.
Since, line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$
Then, $2(1)-3(0)-4(-2)-10=0$
          $\Rightarrow 0=0$
and $(i+2j-k).(2i-3j-4k)=0$
       $\Rightarrow 2-6+4=0$
       $\Rightarrow 0=0$
Therefore, statement 1 is true.

Ans: A

If $\theta$ denotes the acute angle between the line $\bar{r} = (\bar{i} + 2\bar{j} - \bar{k}) + \lambda  (\bar{i} - \bar{j} + \bar{k})$ and the plane $\bar{r} = (2\bar{i} - \bar{j} + \bar{k}) = 4$, then $\sin \theta + \sqrt 2 \cos \theta$

  1. $\dfrac{1}{\sqrt 2}$

  2. $1$

  3. $\sqrt 2$

  4. $1 + \sqrt 2$


Correct Option: A

Let $\vec {AB}=\hat {i}-\hat {j}+\hat {k}$ be rotated about $A$ along the plane $3x-y-2z=5$ by an angle $\cos^{-1}\dfrac {\sqrt {2}}{3}$ so that the point $B$ reaches the point $C$, then the vector representing $AC$ may be

  1. $\dfrac {\sqrt {3}(-2\hat {j}+\hat {k})}{\sqrt {5}}$

  2. $\dfrac {\hat {i}-\hat {j}+2\hat {k}}{\sqrt {2}}$

  3. $\dfrac {\sqrt {3}(\hat {i}+3\hat {j})}{\sqrt {10}}$

  4. $\dfrac {\hat {i}-7\hat {j}+2\hat {k}}{3\sqrt {2}}$


Correct Option: A

Gives the line $\displaystyle L:\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-3 }{ -1 } $ and the plane $\pi :x-2y=0$. Of the following assertions, the only one that is always true is:

  1. $L$ is $\bot$ to $\pi$

  2. $L$ lies in $\pi$

  3. $L$ is parallel to $\pi$

  4. none of these


Correct Option: B
Explanation:

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0$

$\therefore$ given line is $\bot$ to the normal to the plane i.e. given line is parallel to the given plane.
Also, $(1,-1,3)$ lies on the plane $x-2y-z=0$
$1-2\left( -1 \right) -3=0\Rightarrow 1+2-3=0$
which is true
$\therefore L$ lies in plane $\pi$

Consider a plane $x + y - z = 1$ and the point $A(1, 2, -3)$. A line $L$ has the equation $x = 1 + 3r$, $y = 2 - r$, $z = 3 + 4r$

The coordinate of a point $B$ of line $L$, such that $AB$ is parallel to the plane, is

  1. $(10, -1, 15)$

  2. $(-5, 4, -5)$

  3. $(4, 1, 7)$

  4. $(-8, 5, -9)$


Correct Option: D
Explanation:

Let $\vec { OB } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k$
$\vec { AB } =\vec { OB } -\vec { OA } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k-\hat i-2\hat j+3\hat k=3r\hat i-r\hat j+\left( 6+4r \right)\hat k$
Since, $\vec { AB }$ is parallel to $x+y-z=1$
Therefore, $\vec { AB } .\left(\hat i+\hat j-\hat k \right) =0$
$\Rightarrow \left( 3r\hat i-r\hat j+\left( 6+4r \right)\hat k \right) .\left(\hat i+\hat j-\hat k \right)=0 $
$\Rightarrow 3r-r-6-4r=0$
$\Rightarrow r=-3$
Therefore, $\vec { OB } =-8i+5j-9k$

Ans: D

If the angle between the line $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { 5/14 }  \right)  } $ then $\lambda$=

  1. $\dfrac{3}{2}$

  2. $\dfrac{5}{3}$

  3. $\dfrac{2}{3}$

  4. $\dfrac{2}{5}$


Correct Option: A

Consider plane containing line $\dfrac{x+1}{-3} = \dfrac{y-3}{z} = \dfrac{z+2}{-1}$ and passing through the point $(1, -1, 0)$. The angle made by the plane with x-axis is

  1. $tan^{-1} \sqrt{2}$

  2. $tan^{-1} \sqrt{2}$

  3. $\dfrac{\pi}{6}$

  4. none of these


Correct Option: A

Consider plane containing line $\dfrac{x+1}{-3} = \dfrac{y-3}{2} = \dfrac{z+2}{-1}$ and passing through the point $(1, -1, 0)$ . The angle made by the plane with x-axis is 

  1. $tan^{-1} \sqrt{2}$

  2. $cot^{-1} \sqrt{2}$

  3. $\dfrac{\pi}{6}$

  4. none of these


Correct Option: A