Tag: applications of vector algebra

Questions Related to applications of vector algebra

The angle between the line $\dfrac{x-1}{1}=\dfrac{y+2}{1}=\dfrac{z-4}{0}$ and the plane $y+z+2=0$ is

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

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

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

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

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

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Correct Option: A
Explanation:

$\text cos\theta = \dfrac{1.0+1.1+0.1}{(√(1^{2}+1^{2}+0^{2}).(√0^{2}+1^{2}+1^{2})}$

$\text cos\theta = \dfrac{1}{√2.√2}$
$\text cos\theta = \dfrac{1}{2}$
$\theta = \dfrac{π}{3}$

The angle between the line $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$ and the plane $3x + 2y - 3z = 4$, is

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

  1. $45^o$

  2. $0^o$

  3. $\cos^{-1} \left(\dfrac{24}{\sqrt{29 \times 22}}\right)$

  4. $90^o$

Show answer
Correct Option: B
Explanation:

Line$:\cfrac { x }{ 2 } =\cfrac { y }{ 3 } =\cfrac { z }{ 4 } $ has directions $(2,3,4)$

Plane$:3x+2y-3z=4$ has normal with direction ratios $(3,2,-3)$
$\therefore$ angle between plane and line be $\theta$ then angle between line and its direction will be $90-\theta$.
$\cos { (90-\theta ) } =\cfrac { 2\times 3+3\times 2+4\times -3 }{ \sqrt { ({ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 })({ 3 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }) }  } =0\ \therefore 90-\theta =90\quad \Rightarrow \theta ={ 0 }^{ \circ  }$

The projection of the line segment joining the points $(1, 2, 3)$ and $(4, 5, 6)$ on the plane $2x + y + z = 1$ is 

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

  1. $1$

  2. $\sqrt{3}$

  3. $5$

  4. $6$

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Correct Option: B
Explanation:

Points $A(1,2,3)$ and $B(4,5,6)$ ,  Plane :$2x+y+z=1$

length of projection is distance between foot of perpendicular of $A$ & $B$ on Plane.
Directions of normal to plane $\Rightarrow 2,1,1$
let $(2r+1,r+2,r+3)$ is foot of $\bot$ from $A(1,2,3)$.
$(2r+1)2+(r+2)+(r+3)=1\ \Rightarrow r=-1$
foot of $\bot$ from $A=(-1,1,2)$
Similarly,If $(2k+4,k+5,k+6)$ is foot of $\bot$ from $B(4,5,6)$
$\therefore (2k+1)2+(k+5)+(k+6)=1\ \Rightarrow k=-3$
foot of $\bot$ from $B=(-2,2,3)$
$\therefore$ distance between foot of $\bot$ from $A$ & foot of $\bot$ from $B$.
$\Rightarrow \sqrt { { (-2-(-1)) }^{ 2 }+{ (2-1) }^{ 2 }+{ (3-2) }^{ 2 } } =\sqrt { 3 } $

If $\overline {c}$ is perpendicular to $\overline {a}$ and $\overline {b}$ , $\left| \overline {a} \right| =3,\ \left| \overline {b} \right|=4,\ \left| \overline {c} \right|=5$ and the angle between $\overline {a}$ and $\overline {b}$ is $\dfrac{\pi}{6}$ then $[\overline {a}\ \ \ \overline {b}\ \ \ \overline {c}]=$

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

  1. $30\sqrt{3}$

  2. $30$

  3. $15$

  4. $15\sqrt{3}$

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Correct Option: B
Explanation:

We have,

$\begin{matrix} \left[ { \overline { a } \, \, \overline { b } \, \, \overline { c }  } \right] =\overline { c } \times \left( { \overline { a } \times \overline { b }  } \right)  \ =\overline { c } \times \left( { \overline { a } \times \overline { b }  } \right) \cos { 0^{ 0 } }  \ =5\times 3\times 4\times \sin  \frac { \pi  }{ 6 }  \  \end{matrix}$
$ = 5 \times 3 \times 4 \times \frac{1}{2}$
$ = 30$
Then,
Option $B$ is correct answer.

An angle between the plane , $x+y+z=5$ and the line of intersection of the planes, $3x+4y+x-1=0$ and $5x+8y+2z+14=0$

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

  1. $\sin^{-1}(\sqrt{3/17})$

  2. $\cos^{-1}(\sqrt{3/17})$

  3. $\cos^{-1}(3/\sqrt{17})$

  4. $\sin^{-1}(3/\sqrt{17})$

Show answer
Correct Option: A
Explanation:

We have

$\overrightarrow {{\pi _1}} :x + y + z = 5$
$\overrightarrow {{r _1}} :3x + 4y + z - 1 = 0$
$\overrightarrow {{r _2}} :5x + 8y + 2z + 14 = 0$
Line of intersection of planes $\parallel \,\,to\,\,\overrightarrow {{r _1}}  \times \overrightarrow {{r _1}} $
By the helps of determinate
$\left| { \begin{array} { *{ 20 }{ c } }{ \widehat { i }  } & { \widehat { j }  } & { \widehat { k }  } \ 3 & 4 & 1 \ 4 & 8 & 2 \end{array} } \right| $
$ = \widehat i\left( 0 \right) - \widehat j\left( {6 - 5} \right) + \widehat k\left( {24 - 20} \right)$
$ =  - \widehat j + 4\widehat k$
Now,
$\sin \theta  = \frac{{ - 1 + 4}}{{\sqrt 3 \sqrt {17} }} = \sqrt {\frac{3}{{17}}} $
$\therefore \theta  = {\sin ^{ - 1}}\sqrt {\frac{3}{{17}}} $
Hence the option $A$ is the correct answer.

Read the following statement carefully and identify the true statement
(a) Two lines parallel to a third line are parallel
(b) Two lines perpendicular to a third line are parallel
(c) Two lines parallel to a plane are parallel
(d) Two lines perpendicular to a plane are parallel
(e) Two lines either intersect or are parallel

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

  1. a & b

  2. a & d

  3. d & e

  4. a

Show answer
Correct Option: B,D
Explanation:
$(a)$ Two lines parallel to a third line are parallel-True

$(b)$ Two lines perpendicular to a third line are parallel-False. 

The $x-$ and $y-$axes are both perpendicular to the $z-$axis, yet the $x-$ and $y-$axes are not parallel.

$(c)$ Two lines parallel to a plane are parallel-False. 

The $x-$ and $y-$axes are not parallel, yet they are both parallel to the plane $z = 1.$

$(d)$ Two lines perpendicular to a plane are parallel-True

$(e)$ Two lines either intersect or are parallel-False. They can be skew.
Only the statements $(a)$ and $(d)$ are true.

The line $\dfrac {x - 2}{3} = \dfrac {y - 3}{4} = \dfrac {z - 4}{5}$ is parallel to the plane.

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

  1. $3x + 4y + 5z = 7$

  2. $2x + y - 2z=0$

  3. $x + y - z = 2$

  4. $2x + 3y$

Show answer
Correct Option: B
Explanation:

Direction ratios of given line are $3,4,5$ direction ratios of the perpendicular to the  plane $2x+y-2z=0$ are $2,1,-2$


Now
$\begin{array}{l} 2\times 3+1\times 4+\left( { -2 } \right) \times 5 \ =6+4-10 \ =0 \end{array}$

perpendicular to the plane is perpendicular to the given line
so, the plane $2x+y-2z$ is parallel to the given line.

Hence, the correct option is $B$

If the projection of point P$(\vec{p})$ on the plane $\vec{r}\cdot \vec{n}=q$ is the points $S(\vec{s})$, then.

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

  1. $\vec{s}=\dfrac{(q-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

  2. $\vec{s}=\vec{p}+\dfrac{(q-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

  3. $\vec{s}=\vec{p}-\dfrac{(\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

  4. $\vec{s}=\vec{p}-\dfrac{(\vec{q}-\vec{p}\cdot \vec{n})\vec{n}}{|\vec{n}|^2}$

Show answer
Correct Option: B
Explanation:

Consider the problem 

Let,
$\pi : \vec r.\vec n=q$ be the plane.
Now $S$ will lie on $\pi $. 
Consider $\vec S$ will joins $O(0,0,0)$ and  $S$. And as $P$ is projection on plane, 
then 
$\vec {PS}$ is perpendicular to plane.

$\vec P=\vec {OP}$
Now In triangle $OSP$, by triangle law of vector addition.

$\vec {OS}+\vec {SP}=\vec {OP}$  

$|\vec {SP}|=$ distance between $P$ and plane 
$=\dfrac{\vec p.\vec n-q}{|\vec n|}$

$\vec {SP}=|\vec {SP}|.\hat {SP}$

Now, 
$\hat {SP}=\hat n=\dfrac{\vec n}{|\vec n|}$

$\vec {SP}=\dfrac{|\vec {SP}|\vec n}{|\vec n|}$

$\vec {SP}=\dfrac{(\vec p.\vec n-q)\vec n}{|\vec n|^2}$

$\vec {OS}+\vec {OP}=\vec {OP}$
$\vec S+\vec {SP}=\vec P$

$\vec S=\vec p-\vec {SP}$
$=\vec p-(\dfrac{\vec p.\vec n-q}{|\vec n|^2})\vec n$

$\vec S=\vec p+\dfrac{(q-\vec p.\vec n)\vec n}{|\vec n|^2}$

The line $\cfrac{x+3}{3}=\cfrac{y-2}{-2}=\cfrac{z+1}{1}$ and the plane $4x+5y+3z-5=0$ intersect at a point

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

  1. $(3,1,-2)$

  2. $(3,-2,1)$

  3. $(2,-1,3)$

  4. $(-1,-2,-3)$

Show answer
Correct Option: B
Explanation:

Let $\dfrac{x+3}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{1}=k$


On solving, we get

$\Rightarrow x=3k-3$

$\Rightarrow y=-2k+2$ 

$\Rightarrow z=k-1$

On substituing these values in the given plane equation we get,

$\Rightarrow 4x+5y+3z-5=0$

$\Rightarrow 4(3k-3)+5(-2k+2)+3(k-1)-5=0$

On simpliying we get,

$\Rightarrow 5k=10$

$\Rightarrow k=2$

Substituting this value of $k$ in equations of $x,y,z$ we get

$\Rightarrow x=3,y=-2,z=1$

Hence point of intersection is $(3,-2,1)$

If $a,b$ and $c$ are three unit vectors equally inclined to each other at angle $\theta$. Then, angle between $a$ and the plane of $b$ and $c$ is

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

  1. $\cos ^{ -1 }{ \left( \cfrac { \cos { \theta } }{ \cos { \left( \theta /2 \right) } } \right) } $

  2. $\sin ^{ -1 }{ \left( \cfrac { \sin { \theta } }{ \sin { \left( \theta /2 \right) } } \right) } $

  3. $\sin ^{ -1 }{ \left( \cfrac { \cos { \theta } }{ \cos { \left( \theta /2 \right) } } \right) } $

  4. $\cos ^{ -1 }{ \left( \cfrac { \sin { \theta } }{ \sin { \left( \theta /2 \right) } } \right) } $

Show answer
Correct Option: A