Tag: applications of vector algebra

Questions Related to applications of vector algebra

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

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\cfrac{15}{2}$

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

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

  4. $\cfrac{5}{3}$

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Correct Option: A
Explanation:

equation of line $\dfrac{x-0}{1}=\dfrac{y-1}{2}=\dfrac{2-3}{\lambda}$ where $a _1=1, a _2=2, a _3=\lambda$

equation of plane $n+2y+32=4$
where $b _1=1, b _2=2, b _3=3$
angle between Line and plane
$\cos\theta=\dfrac{|a _1b _1+a _2b _2+a _3b _3|}{\sqrt{a _1^2+a _2^2+a _3^2}.\sqrt{b _1^2+b _2^2+b _3^2}}$
$\cos \theta=\dfrac{|1.1+2.2+3.\lambda|}{\sqrt{1+4+\lambda^2}.\sqrt{1+4+9}}$
$\cos \theta =\dfrac{5+3\lambda}{\sqrt{5+\lambda^2}.\sqrt{14}}\quad ---(1)$
given $\theta =\cos^{-1}\left(\sqrt{\dfrac{5}{14}}\right)$
$\cos \theta =\left(\sqrt{\dfrac{5}{14}}\right)\quad ----(2)$
By eqn $(1)$ & $(2)$
$\dfrac{5+3\lambda }{\sqrt{5+\lambda^2}\sqrt{14}}=\sqrt{\dfrac{5}{14}}$
$5+3\lambda=\sqrt{5}.\sqrt{5+\lambda^2}$
$(5+3\lambda)^2=5.(5+\lambda^2)$
$25+9\lambda^2+30\lambda =25+5\lambda^2$
$4\lambda^2+30\lambda =0$
$\lambda (2\lambda +15)=0\Rightarrow \lambda =0, \dfrac{15}{2}$
but $\lambda \neq 0$
So $\lambda =\dfrac{15}{2}$ Ans

How is the line $\displaystyle \frac{x-4}{4}=\frac{y-12}{12}=\frac{z-8}{8}$ related to the planes
(A) $\displaystyle x-y+z=0$
(B) $\displaystyle x-y+z-6=0$

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. parallel to plane A but not B

  2. parallel to plane A and also lies in plane A but not parallel to B

  3. parallel to plane A and also lies in plane A

  4. none of these

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Correct Option: C
Explanation:

A line is inclined at Φ to a plane. The vector equation of the line is given by 

$\vec r =\vec a + \lambda \vec b $
Let $\theta$ be the angle between the line and the normal to the plane. Its value can be given by the following equation
$cos\theta=|\dfrac{\vec b . \vec n }{|\vec b|.|\vec n |}|$
Finding the value of the $Φ$ between the line and the plane we know that 
parallel to plane A and also lies in plane A.

If the angle $\theta $ between the line $\displaystyle \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda} z+4=0$ is such that $\displaystyle \sin \theta =\frac{1}{3}$, then value of $\lambda $ is

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\displaystyle -\frac{3}{5}$

  2. $\displaystyle \frac{5}{3}$

  3. $\displaystyle -\frac{4}{3}$

  4. $\displaystyle \frac{3}{4}$

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Correct Option: B
Explanation:

Angle between the line and plane is same as the angle between the line and normal to the plane
$\displaystyle \therefore \cos \left ( 90^{0}-\theta  \right )=\frac{a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}}{\sqrt{a _{1}^{2}+b _{1}^{2}+c _{1}^{2}}\sqrt{a _{2}^{2}+b _{2}^{2}+c _{2}^{2}}}$
$\displaystyle \therefore\frac{1}{3}=\frac{\left ( 1\times 2+2\times \left ( -1 \right )+2\sqrt{\lambda } \right )}{\sqrt{1^{2}+2^{2}+2^{2}}\sqrt{2^{2}+1^{2}+\lambda }} $

$\therefore  \lambda =\dfrac{5}{3}$

If $\displaystyle \theta$ is the angle between the line 
$\vec r=2i+j-k+\left ( i+j+k \right )t$ and the plane
$\displaystyle \vec r\cdot \left ( 3i-4j+5k \right )=q$, then

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\displaystyle \cos \theta =\frac{2\sqrt{6}}{15}$

  2. $\displaystyle \sin \theta =\frac{2\sqrt{6}}{15}$

  3. $\displaystyle \sin \theta =-\frac{11\sqrt{7}}{70}$

  4. $\displaystyle \cos \theta =-\frac{11\sqrt{7}}{70}$

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Correct Option: B
Explanation:

 $\theta$ is angle b/w $\xrightarrow [\gamma]{} =2\hat {  i}+j+k+(i+j+k)t$ and $\rightarrow.(3\hat { i }-4\hat { j }+5k)=q$

Angle b/w line and plane is given by 

$\sin\theta =\dfrac{4 _1a _2+b _1b _2+c _1c _2}{\sqrt{a _1^2+b _1^2+c _1^2}\sqrt{a _2^2+b _2^2+c _2^2}}$   

Where $(a _1,b _1,c _1)$ and $(a _2,b _2,c _2)$ are direction ratios of line and plane Respectively so here 

$a _1,b _1,c _1)=(1,1,1)$ and $(a _2,b _2,c _2)=(3,-4,5)$

So $\sin \theta=\dfrac{3-4+5}{\sqrt{1+1+1}\sqrt{9+16+25}}$

$\dfrac{4}{\sqrt{3}\sqrt{50}}=\dfrac{4}{\sqrt{3}5\sqrt{2}}=\dfrac{4}{\sqrt{6.5}}\times \dfrac{\sqrt{6}}{\sqrt{6}}=\dfrac{2\sqrt{6}}{5.3}=\dfrac{2\sqrt{6}}{15}$

so here $\sin\theta =\dfrac{2\sqrt{6}}{15} \Rightarrow \theta =\sin\dfrac{2\sqrt{6}}{15}$

The projection of line $\displaystyle\frac{x}{2}=\frac{y-1}{2}=\frac{z-1}{1}$ on a plane 'P' is $\displaystyle\frac{x}{1}=\frac{y-1}{1}=\frac{z-1}{-1}$. If the plane P passes through $(k, -2, 0)$, then k is greater than.

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $2$

  2. $3$

  3. $5$

  4. $4$

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Correct Option: A