MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 12

$P$ ' is a point inside the triangle

$ABC$ , such that

$\displaystyle BC\left ( \vec{PA} \right )+CA\left ( \vec{PB} \right )+AB\left ( \vec{PC} \right )=0,$ then for the triangle

$ABC$ the point

$P$ is its :

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Incentre
0%
Circumcentre
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Centroid
0%
Orthocentre

Explanation

Given that,

$P$ is a point inside the triangle

$ABC$ , such that

$\displaystyle BC\left ( \vec{PA} \right )+CA\left ( \vec{PB} \right )+AB\left ( \vec{PC} \right )=0$

Let,

$BC=a,CA=b,AB=c$ and

$\vec{OA}=\bar{a},\vec{OB}=\bar{b},\vec{OC}=\bar{c}$

$\Rightarrow a(\bar{a}-\vec{OP})+b(\bar{b}-\vec{OP})+c(\bar{c}-\vec{OP})=0$

$\Rightarrow \vec{OP}=\displaystyle\dfrac{a\bar{a}+b\bar{b}+c\bar{c}}{a+b+c}$

$\therefore P$ is an Incentre of triangle

$ABC$ .

Hence, option

$A$ .

Let the pairs

$\vec{a}, \vec{b}$ and

$\vec{c}, \vec{d}$ each determine a plane, then the planes are parallel if

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$(\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) = \vec{0}$
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$(\vec{a} \times \vec{c}) \cdot (\vec{b} \times \vec{d}) = \vec{0}$
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$(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$
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$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \vec{0}$

Explanation

$\vec{a} \times \vec{b}$ is a vector perpendicular to the plane contining

$\vec{a}$ and

$\vec{b}$ .

Similarly,

$\vec{c} \times \vec{d}$ is a vector perpendicular to the plane containing

$\vec{c}$ and

$\vec{d}$ .

Thus, the two planes will be parallel if their normals,
i.e.,

$\vec{a} \times \vec{b}$ and

$\vec{c} \times \vec{d}$ , are parallel.

Thus,

$(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$

Hence, option

$C$ .

Let

$\vec{a},\vec{b},\vec{c}$ be vectors of length

$3,4,5$ respectively. Let

$\vec{a}$ be perpendicular to

$\vec{b}+\vec{c},\vec{b}\,to\,\vec{c}+\vec{a}$ and

$\vec{c}\,to\,\vec{a}+\vec{b}$ . Then

$\begin{vmatrix}\vec{a}+\vec{b}+\vec{c}\end{vmatrix}$ is

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$\;2\sqrt{5}$
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$\;2\sqrt{2}$
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$\;10\sqrt{5}$
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$\;5\sqrt{2}$

Explanation

$\begin{vmatrix}\vec{a}+\vec{b}+\vec{c}\end{vmatrix}=\sqrt{(\vec{a}+\vec{b}+\vec{c})^2} =\sqrt{a^2+b^2+c^2+2\begin{pmatrix}\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}\end{pmatrix}}$ {Modulus formula for vectors}

$\because\,\vec{a}.\begin{pmatrix}\vec{b}+\vec{c}\end{pmatrix}=0,\;\vec{b}.\begin{pmatrix}\vec{c}+\vec{a}\end{pmatrix}=0\,\&\,\vec{c}.\begin{pmatrix}\vec{a}+\vec{b}\end{pmatrix}=0$ [As the given conditions of being perpendicular}

$\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=0$ {expanding the previous expression and substituting in the first expression}

$\Rightarrow \begin{vmatrix}\vec{a}+\vec{b}+\vec{c}\end{vmatrix}=\sqrt{a^2+b^2+c^2}=\sqrt{3^2+4^2+5^2}=5\sqrt{2}$

$ABCDEF$ is a regular hexagon . The centre of hexagon is a point O. Then the value of

$\overrightarrow{AB}+ \overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}$ is

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$2\overrightarrow{AO}$
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$4\overrightarrow{AO}$
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$6\overrightarrow{AO}$
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Zero

Explanation

We have

$\overrightarrow{AB}+ \overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}$
We have

$\overrightarrow{AB}+ \overrightarrow{AC}=\overrightarrow{AB}+ (\overrightarrow{AB}+\overrightarrow{BC})=2\overrightarrow{AB}+\overrightarrow{BC}.............(i)$
and

$\overrightarrow{AE}+\overrightarrow{AF}=(\overrightarrow{AD}+\overrightarrow{DE})+(\overrightarrow{AD}+\overrightarrow{DE}+\overrightarrow{EF})=2\overrightarrow{AD}+2\overrightarrow{DE}+\overrightarrow{DF}............(ii)$
From

$i$ and

$ii$ we see that

$\overrightarrow{AB}$ and

$\overrightarrow{AC}$ are vectorially opposite to

$\overrightarrow{DE}$ and

$\overrightarrow{DF}$
Thus we get

$\overrightarrow{AB}+ \overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}= 3\overrightarrow{AD}=6\overrightarrow{AO}$

The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors a, b, c such that a.b = b.c = c.a = 1/What is the volume of the parallelopipe.

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$\dfrac{1}{\sqrt{2}}$
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$\dfrac{1}{\sqrt{3}}$
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$\dfrac{3}{\sqrt{2}}$
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None of these

If the sum of two unit vectors is also a unit vector, then the angle between the two vectors is

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$\dfrac{\pi}{3}$
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$\dfrac{2\pi}{3}$
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$\dfrac{\pi}{4}$
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None of these

Explanation

Given condition

$\vec { a } +\vec { b } =\vec { c }$

where

$\left| \vec { a } \right| =\left| \vec { b } \right| =\left| \vec { c } \right| =1$

Squaring both sides

${ \left| \vec { a } \right| }^{ 2 }+{ \left| \vec { b } \right| }^{ 2 }+2.\left| \vec { a } \right| .\left| \vec { b } \right| .\cos { \theta } ={ \left| \vec { c } \right| }^{ 2 }$

$1+1+2\cos { \theta } ={ \left| \vec { c } \right| }^{ 2 }=1$

$2\cos { \theta } =-1$

$\cos { \theta } =\cfrac { -1 }{ 2 }$

$\cos { \theta } =\cos { \left( \pi -\cfrac { \pi }{ 3 } \right) }$

$\theta =\cfrac { 2\pi }{ 3 }$

Let

$b = 4i + 3j$ and

$c$ be two vectors perpendicular to each other in the xy-plane. If

$r_i, \ i =1, 2 ... n$ , are the vectors in the same plane having projections

$1$ and

$2$ along

$b$ and

$c$ respectively then

$\displaystyle \sum_{i=1}^{n} \left| r_{i}\right|^{2}$ is equal to

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$20$
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$10$
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$4$
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$7$

Explanation

Let

$r=\lambda b + \mu c$ and

$c = \pm (xi + yj).$ Since

$c$ and

$b$ are perpendicular, we have

$\displaystyle 4x + 3y = 0 \Rightarrow c= \pm x\left( i-\frac{4}{3}j\right)$

$\pm 1= proj\cdot of$ r

$on$ b

$\displaystyle=\frac{ r\cdot b}{\left| b\right| }=\frac{(\lambda b+\mu c)\cdot b}{\left| b\right| } =\frac{\lambda b\cdot b}{\left| b\right| } [\because b\cdot c=0]$

$\lambda \left| b\right| =5 \lambda$ Hence

$\lambda=\pm 1/5$
Also

$\pm 2=proj.$ of

$r$ on

$\displaystyle c=\frac {r\cdot c}{\left|c\right|}$

$\displaystyle =\frac{(\lambda b+\mu c)\cdot c}{\left| c\right|} =\mu \left| c\right| =\frac{5}{3} \mu x$
Thus,

$\mu x =\pm 6/5$ Therefore,

$\displaystyle r=\frac{1}{5} (4i+3j)+\frac{6}{5}\left( i-\frac{4}{3}j\right) =\pm (2i-j,)$

$\displaystyle r=\frac{1}{5} (4i+3j)-\frac{6}{5}\left( i-\frac{4}{3}j\right) =\pm \left (-\frac{2}{5} i+ \frac{11}{5}j,\right)$
Thus there are four such vectors

$\displaystyle \sum_{i=1}^{4} \left| r_{i}\right|^{2}=2 \left| 2i -j\right| ^{2} +\left| -\frac{ 2}{5} i + \frac{11}{5} j\right|^{2} =20$

The magnitude of the projection of the vector

$\overline{a} =4\overline{i}-3\overline{j}+2\overline{k}$ on the line which makes equal angles with the coordinate axes is

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

Explanation

Given

$a=4i-3j+2k$

The line which makes equal angle with the coordinate axis is

$b=i+j+k$

The projection of

$a$ on

$b$ is

$\dfrac{a.b}{b} = \dfrac{4-3+2}{\sqrt3}=\dfrac{3}{\sqrt3}=\sqrt3$

Let

$x_{0}$ and

$x_{1}$ be the critical points of

$\displaystyle f(x) =\int_{1}^{x}(t(t + 1) (t + 2) (t + 3)- 24).dt$ and

$\vec r$ &

$\vec r'$ be the parallel vectors with

$\left| \vec r \right| =\left| x_{0}\right|$ and

$\left| \vec r\ ' \right| =\left|x_{1}\right| ,$ then

$\vec r\cdot \vec r\ '$ is equal to

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$24$
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$16$
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$8$
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$4$

Explanation

$\displaystyle f(x)=\int _{ 1 }^{ x } (t(t+1)(t+2)(t+3)24)dt$

$f'( x) =x(x + 1)(x + 2)(x + 3) -24$ ,

Clearly

$f'(1) =0,$
Dividing

$f'(x)$ by

$x -1$ , we get

$f'(x) = (x -1)(x + 4)(x^{2} + 3x + 6)$
Thus the critical points are

$x = 1, -4.$

Hence

$r \cdot r'= \left| x_{0}\right| \left| x_{1}\right| =1\times 4 = 4.$

If one point on the vector

$2i -4j-k$ is

$(2,1,3)$ ,the other point is?

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

Explanation

Let the other point on the vector

$2i-4j-k$ be

$(x,y,z)$ . Given point on the vector is

$(2,1,3)$ .

Thus the vector is represented as

$(2,-4,-1) = (x,y,z)-(2,1,3)$

$(2,-4,-1)=(x-2,y-1,z-3)$

Equating the corresponding points we get

$2=x-2, -4=y-1, -1=z-3$

$\Rightarrow x=4, y=-3, z=2$

Hence the other point on the vector is

$(4,-3,2)$ .

If the vectors

$\overline { c } ,\overline { a } =x\hat { i } +y\hat { j } +z\hat { k }$ , and

$\overline { b } =\hat { j }$ are such that

$\overline { a } ,\overline { c } \ and \ \overline { b }$ from a right handed system, then

$\overline { c }$ is

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$z\hat { i-x } \hat { k }$
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$\overline { 0 }$
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$y\hat { j }$
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$-z\hat { i+x } \hat { k }$

A point

$C = \dfrac {5\overline {a} + 4\overline {b} - 5\overline {c}}{3}$ divides the line joining the points

$A$ and

$B = 2\overline {a} + 3\overline {b} - 4\overline {c}$ in the ratio

$2 : 1$ , then the position vector of

$A$ is

0%
$\overline {a} + 3\overline {b} - 4\overline {c}$
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$2\overline {a} - 3\overline {b} + 4\overline {c}$
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$2\overline {a} + 3\overline {b} + 4\overline {c}$
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$\overline {a} - 2\overline {b} + 3\overline {c}$

$\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$ ,

$\overrightarrow{b} = \hat{i} + 2\hat{j} - \hat{k}$ ,

$\overrightarrow{c} = \hat{i} + \hat{j} - 2\hat{k}$ , then a vector in the plane of

$\hat{b}$ and

$\hat{c}$ whose projection on

$\hat{a}$ is a magnitude of

$\sqrt{\frac{2}{3}}$ is

0%
$2\hat{i} + 3\hat{j} - 3\hat{k}$
0%
$2\hat{i} + 3\hat{j} + 3\hat{k}$
0%
$-2\hat{i} - \hat{j} + 5\hat{k}$
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$2\hat{i} + \hat{j} + 5\hat{k}$

If P(x, y, z) is a point on the line segment joining Q(2, 2, 4) and R(3, 5, 6) such that the projections of

$\overrightarrow{OP}$ on the axes are

$\dfrac{13}{5}, \dfrac{19}{5}, \dfrac{26}{5}$ respectively, then P divides QR in the ratio

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1 : 2
0%
3 : 2
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2 : 3
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1 : 3

Explanation

Let

$P \left ( x,y,z \right )$

$Q \left ( 2,2,4 \right )$

$R \left ( 3,5,6 \right )$

Let the ratio in which

$P \left ( x,y,z \right )$ divides the line joining

$Q \left ( 2,2,4 \right ), R \left ( 3,5,6 \right )$ be

$\lambda :1$

According to the section formula ,

$P\left ( x,y,z \right ) =\left ( \dfrac{3\lambda +2}{\lambda +1} ,\dfrac{5\lambda +2}{\lambda +1},\dfrac{6\lambda +4}{\lambda +1}\right )$

Projection of

$\vec{a}$ on

$\vec{b}$ is

$\dfrac{\vec{a}\cdot \vec{b}}{ \left | \vec{b} \right |}$

It is given that projection of

$\vec{OP}$ on coordinate axes are

$\dfrac {13}{5}, \dfrac {19}{5}, \dfrac{26}{5}$ is

$\dfrac{\left ( x,y,z \right )\cdot \left ( 1,0,0 \right )}{ \left | 1,0,0 \right |} = \dfrac{13}{5}$

$x= \dfrac{13}{5} = \dfrac{3\lambda +2}{\lambda +1}$

$\lambda =\dfrac{3}{2}$

What are coinitial vectors.?

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Two or more vectors having the same magnitude are called coinitial vectors.
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Two or more vectors are said to be coinitial if they are parallel to the same line, irrespective of their magnitudes and directions.
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Two or more vectors having the same initial point are called coinitial vectors.
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None of the above

Explanation

two vectors having same magnitude called equal vector

Hence option

$A$ is not correct

option $$B$4 also is the property of equal vector

coinitail means the initail point is same

So option

$C$ is correct

If three non-zero vectors are

$a=a_1i+a_2j+a_3k , \, b=b_1i+b_2j+b_3k$ and

$c=c_1i+c_2j+c_3k$ . If c is the unit vector perpendicular to the vectors a and b the angle between a and b is

$\dfrac{\pi}{6}$ , then

$\begin{vmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}$ is equal to

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0
0%
$\dfrac{3(\sum a^2_1)(\sum b^2_1) \sum c^2_1}{4}$
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1
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$\dfrac{(\sum a^2_1)(\sum b^2_1)}{4}$

A unit vector a makes an angel

$\Pi /4$ with the z-axis. If a+i+j is a unit vector, then a can be equal to

0%
$\dfrac { i }{ 2 } -\dfrac { j }{ 2 } +\dfrac { k }{ \sqrt { 2 } }$
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$\dfrac { i }{ 2 } -\dfrac { j }{ 2 } -\dfrac { k }{ \sqrt { 2 } }$
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$-\dfrac { i }{ 2 } -\dfrac { j }{ 2 } +\dfrac { k }{ \sqrt { 2 } }$
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None

Explanation

Equation of z-axis =

$=0i+0j+\hat { k }$

a is unit vector at angel

$\Pi /4$

$\bar { a } .\bar { z } =\left| \bar { a } \right| \left| \bar { z } \right| \cos { \dfrac { \Pi }{ 4 } }$

Let

$a=xi+y\lambda +2\hat { k }$

$z=\cos { \dfrac { \Pi }{ 4 } } \Rightarrow z=\frac { 1 }{ \sqrt { 2 } }$

$u=xi+yj+zk\quad =\quad \delta i+yj+\dfrac { 1 }{ \sqrt { 2 } } k$

${ x }^{ 2 }+{ y }^{ 2 }=\dfrac { 1 }{ 2 }$

$a+i+j$ is unit vector

$\left( x+1 \right) i+\left( y+1 \right) j+\dfrac { 1 }{ \sqrt { 2 } } k$ is unit vector

${ (x+1 })^{ 2 }+{ (y+1) }^{ 2 }=\dfrac { 1 }{ 2 }$

${ (x+1 })^{ 2 }+{ (y+1) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }$

${ x }^{ 2 }={ (x+1 })^{ 2 }=1\quad \quad -x=x+1\\ a=-\dfrac { 1 }{ 2 } i-\dfrac { 1 }{ 2 } j+\dfrac { 1 }{ \sqrt { 2 } } k\quad \quad x=-\dfrac { 1 }{ 2 }\quad Similarly \quad y=-\dfrac { 1 }{ 2 }$

Let

$\vec{u},\ \vec{v},\ \vec{w}$ be such that

$\left| \vec { u } \right| =1$ ,

$\left| \vec { v } \right| =2$ ,

$\left| \vec { w } \right| =3$ . If the projection of

$\vec{v}$ along

$\vec{u}$ is equal to projection of

$\vec{w}$ along

$\vec{u}$ and

$\vec{v}$ and

$\vec{w}$ are perpendicular to each other then

$\left| \bar { u } -\bar { v } +\bar { w } \right|$ equals-

0%
$2$
0%
$\sqrt{7}$
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$\sqrt{14}$
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$14$

Let

$\vec { p }$ and

$\vec { q }$ be the position vectors of the points

$P$ and

$Q$ respectively with respect to origin

$O$ . The points

$R$ and

$S$ divide

$PQ$ internally and externally respectively in the ratio

$2:3$ . If

$\overrightarrow { OR }$ and

$\overrightarrow { OS }$ are perpendicular, then which one of the following is correct?

0%
$9{ p }^{ 2 }=4{ q }^{ 2 }$
0%
$4{ p }^{ 2 }=4{ 9q }^{ 2 }$
0%
$9p=4q$
0%
$4p=9q$

Explanation

$\vec{OR}=\dfrac{3p+2q}{3+2}=\dfrac{1}{5}(3p+2q)$

Given that

$\vec{OR}$ and

$\vec{OS}$ are perpendicular.

$\Rightarrow \vec{OR} \cdot \vec{OS}=0$

$\Rightarrow \dfrac{1}{5}(3p+2q) \cdot (3p-2q)=0$

$\Rightarrow 9|p^2|-4|q^2|=0$

$\Rightarrow 9p^2=4q^2$

Let

$O$ be an interior point of

$\triangle ABC$ such that

$\overline {OA} + 2\overline {AB} + 3\overline {OC} = \overline {O}$ , then the ratio of

$\triangle ABC$ to area of

$\triangle AOC$ is

0%
$2$
0%
$\dfrac {3}{2}$
0%
$3$
0%
$\dfrac {5}{2}$

Explanation

Refer to the figure. Let O be the origin.

Let

$\bar { OA } =\bar { a }$

$\bar { OB } =\bar { b }$

$\bar { OC } =\bar { c }$

Given,

$\bar { OA } +2\bar { OB } +3\bar { OC } =0$

$\therefore \bar { a } +2\bar { b } +3\bar { c } =0$

$\therefore \bar { a } =-2\bar { b } -3\bar { c }$ (1)

1) Area of triangle ABC is given by,

$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \bar { AB } \times \bar { BC } \right|$

$\therefore A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } -\bar { a } \right) \times \left( \bar { c } -\bar { b } \right) \right|$

From equation (1),

$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } -\left( -2\bar { b } -3\bar { c } \right) \right) \times \left( \bar { c } -\bar { b } \right) \right|$

$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } +2\bar { b } +3\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right|$

$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( 3\bar { b } +3\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right|$

$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } +\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right|$

$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) -\left( \bar { b } \times \bar { b } \right) +\left( \bar { c } \times \bar { c } \right) -\left( \bar { c } \times \bar { b } \right) \right|$

But,

$\bar { b } \times \bar { b } =0$ and

$\bar { c } \times \bar { c } =0$

$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) -\left( \bar { c } \times \bar { b } \right) \right|$

$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) +\left( \bar { b } \times \bar { c } \right) \right|$

$\left( \because \left( \bar { a } \times \bar { b } \right) =-\left( \bar { b } \times \bar { a } \right) \right)$

$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| 2\left( \bar { b } \times \bar { c } \right) \right|$

$\therefore A\left( \triangle ABC \right) =3\left| \left( \bar { b } \times \bar { c } \right) \right|$ (2)

2) Area of triangle AOC is given by,

$\therefore \left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \bar { OA } \times \bar { OC } \right|$

$\therefore \left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \bar { a } \times \bar { c } \right|$

From equation (1),

$A\left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \left( -2\bar { b } -3\bar { c } \right) \times \bar { c } \right|$

$\therefore A\left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| -2\left( \bar { b } \times \bar { c } \right) -3\left( \bar { c } \times \bar { c } \right) \right|$

$\therefore A\left( \triangle AOC \right) =\frac { 2 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) \right|$

$\therefore A\left( \triangle AOC \right) =\left| \left( \bar { b } \times \bar { c } \right) \right|$ (3)

Now,

$\frac { A\left( \triangle ABC \right) }{ A\left( \triangle AOC \right) } =\frac { 3\left| \left( \bar { b } \times \bar { c } \right) \right| }{ \left| \left( \bar { b } \times \bar { c } \right) \right| }$

$\frac { A\left( \triangle ABC \right) }{ A\left( \triangle AOC \right) } =3$

The set of values of

$'c'$ for which the angle between the vectors

$(cx\hat{i}-6\hat{j}+3\hat{k})$ and

$(x\hat{i}-2\hat{j}+2cx\hat{k})$ is acute for every

$x\in R$ is

0%
$(0,\ 4/3]$
0%
$[0,\ 4/3)$
0%
$(11/9,\ 4/3)$
0%
$[0,\ 4/3]$

$\vec { a } ,\vec { b }\ and\ \vec { c }$ unit vector satisfying

$\left| \vec { a } -\vec { b } \right| +\left| \vec { b } -\vec { c } \right| +\left| \vec { c } -\vec { a } \right| =9$ , then

$\left| 2\vec { a } +7\vec { b } +7\vec { c } \right|$ is equal to

0%
2
0%
3
0%
4
0%
5

Let

$\vec {a}, \vec {b}$ and

$\vec {c}$ be vectors forming right hand triad. Let

$\vec {p} = \dfrac {\vec {b} \times \vec {c}}{[\vec {a} \vec {b} \vec {c}]}, \vec {q} = \dfrac {\vec {c} \times \vec {c}}{[\vec {a} \vec {b} \vec {c}]}$ and

$\vec {r} = \dfrac {\vec {a}\times \vec {b}}{[\vec {a}\vec {b}\vec {c}]}$ if

$x\epsilon R^{+}$ then

0%
$x[\vec {a}\vec {b}\vec {c}] + \dfrac {[\vec {p}\vec {q}\vec {r}]}{x}$ has least value $2$
0%
$x^{4} [\vec {a}\vec {b}\vec {c}]^{2} + \dfrac {[\vec {p}\vec {q}\vec {r}]}{x^{2}}$ has least value $(3/2^{2/3})$
0%
$[\vec {p}\vec {q}\vec {r}] > 0$
0%
$x[\vec {p}\vec {q}\vec {r}]$ has maximum value $7$

A vector

$\vec{a}$ has components

$2$ p and

$1$ with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new system,

$\vec{a}$ components

$p+1$ and

$1$ , then?

0%
$p=0$
0%
$p=1$ or $p=-\dfrac{1}{3}$
0%
$p=-1$ or $p=\dfrac{1}{3}$
0%
$p=1$ or $p=-1$

Explanation

$\textbf{Step-1: Find the magnitude of the given vector & simplify.}$

$\text{We know that,}$

$\text{The magnitude of the vector must remain same irrespective of the co-ordinate system.}$

$\implies\quad \left| \overline { a } \right| =\left| { \overline { { a }^{ ' } } } \right|$

$\left| 2p\overset { \wedge }{ i } +\overset { \wedge }{ j } \right| =\left| (p+1)\overset { \wedge }{ { i }^{ ' } } +\overset { \wedge }{ { j }^{ ' } } \right|$

$\implies\quad \sqrt { { (2p) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { { (p+1) }^{ 2 }+{ 1 }^{ 2 } }$

$\textbf{Step-2: Simplify the above results to get the required unknown.}$

$\text{After squaring on both sides we get,}$

$\implies\quad { 4p }^{ 2 }+1={ p }^{ 2 }+1+2p+1$

$\implies\quad { 4p }^{ 2 }+1={ p }^{ 2 }+2p+2$

$\implies\quad { 3 }p^{ 2 }-2p-1=0$

$\implies\quad { 3p }^{ 2 }-3p+p-1=0$

$\implies\quad 3p(p-1)+1(p-1)=0$

$\implies\quad (p-1)(3p+1)=0$

$\implies\quad p=1,\cfrac { -1 }{ 3 }$

$\textbf{Hence, option - B is the correct answer.}$

The vector

$(\hat {i}\times \vec {a}.\vec {b})\hat {i} + (\hat {j} \times \vec {a}.\vec {b})\hat {j} + (\hat {k} \times \vec {a} . \vec {b})\hat {k}$ is equal to

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$\vec {b}\times \vec {a}$
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$\vec {a}$
0%
$\vec {a}\times \vec {b}$
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$\vec {b}$

$ABCDE$ is a pentagon, then

$\vec {AB}+\vec {AE}+\vec {BC}+\vec {DC}+\vec {ED}+\vec {AC}$ equals

0%
$3 \vec {AD}$
0%
$3 \vec {AC}$
0%
$3 \vec {BE}$
0%
$3 \vec {CE}$

Explanation

In the pentagon ABCDE, join AC and AD.

By triangle law,

In $\Delta ABC$ ,

$\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}$

In $\Delta ACD$ ,

$\overrightarrow {AD} + \overrightarrow {DC} = \overrightarrow {AC}$

In $\Delta AED$ ,

$\overrightarrow {AE} + \overrightarrow {ED} = \overrightarrow {AD}$

Now,

$\overrightarrow {AB} + \overrightarrow {AE} + \overrightarrow {BC} + \overrightarrow {DC} + \overrightarrow {ED} + \overrightarrow {AC} = \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right) + \left( {\overrightarrow {AE} + \overrightarrow {ED} } \right) + \overrightarrow {DC} + \overrightarrow {AC}$

$= \overrightarrow {AC} + \left( {\overrightarrow {AD} + \overrightarrow {DC} } \right) + \overrightarrow {AC}$

$= \overrightarrow {AC} + \overrightarrow {AC} + \overrightarrow {AC}$

$= 3\overrightarrow {AC}$

991105_1057198_ans_124f8bc5604a4d0c8b1186cff255b9d8.png

The x-y plane divides the line joining the points

$(-1, 3, 4)$ and

$(2, -5, 6)$ :

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internally in the ratio $2:3$
0%
externally in the ratio $2:3$
0%
internally in the ratio $3:2$
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externally in the ratio $3:2$

Explanation

Let $xy$ plane divides the line joining the points $A\left( { - 1,3,4} \right)$ and $B\left( {2, - 5,6} \right)$ in the ratio $k:1$ .

$x = \cfrac{{2k - 1}}{{k + 1}}$

$y = \cfrac{{ - 5k + 3}}{{k + 1}}$

$z = \cfrac{{6k - 4}}{{k + 1}}$

Since the plane is $xy$ plane, so $z = 0$ ,

$\cfrac{{6k - 4}}{{k + 1}} = 0$

$6k - 4 = 0$

$6k = 4$

$k = \cfrac{4}{6}$

$k = \cfrac{2}{3}$

Therefore, the ratio is $2:3$ internally.

$\vec {a},\vec {b}$ and

$\vec {c}$ are those mutually perpendicular vectors, then the projection of the vector

$\left( l \dfrac{\bar {a}}{|\bar {a}|}+m\dfrac{\bar {b}}{|\bar {b}|}+n\dfrac{(\bar {a}\times \bar {b})}{|\bar {a} \times \bar {b}|}\right)$ along bisector of vectors

$\vec {a}$ and

$\vec {a}$ may be given as ?

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$\dfrac{l^{2}+m^{2}}{\sqrt{l^{2}+m^{2}+n^{2}}}$
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$\sqrt{l^{2}+m^{2}+n^{2}}$
0%
$\dfrac{\sqrt{l^{2}+m^{2}}}{\sqrt{l^{2}+m^{2}+n^{2}}}$
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$\dfrac{l+m}{\sqrt{2}}$

Explanation

Question :- If

$\vec{a},\vec{b}$ and

$\vec{c}$ are three mutually perpendicular

vectors, then projection of vector

$\left ( l\dfrac{\vec{a}}{\left | \vec{a} \right |}+m\dfrac{\vec{b}}{\left | \vec{b} \right |}+n\dfrac{(\vec{a}\times \vec{b})}{\left | \vec{a}\times \vec{b} \right |} \right )$

along bisector of sectors

$\vec{a}$ and

$\vec{b}$ may be given as?

Solution:-

We know for mutually perp. vectors.

$\vec{a}\times \vec{b}=\vec{c}$

$\vec{b}\times \vec{c}=\vec{a}$

$\vec{c}\times \vec{a}=\vec{b}$

$\vec{a}.\vec{b}=\vec{b}.\vec{c}=\vec{c}.\vec{a}=0$ and

$\left | \hat{a} \right |^{2}=1$

Now, A vector parallel to bisector of angle between

vectors

$\vec{a}$ and

$\vec{b}$ is given as

$\vec{x}$

$\vec{x}=\dfrac{\vec{a}}{\left | \vec{a} \right |}+\dfrac{\vec{b}}{\left | \vec{b} \right |}=\hat{a}+\hat{b}$

$(\hat{a}=\hat{b}$ = unit vectors)

say

$\vec{y}=\left ( l\dfrac{\vec{a}}{\left | \vec{a} \right |} +m\dfrac{\vec{b}}{\left | \vec{b} \right |}+n\dfrac{(\vec{a}\times \vec{b})}{\left | \vec{a}\times \vec{b} \right |}\right )$

$\vec{y}=\left ( l\hat{a}+m\hat{b}+n\dfrac{\vec{c}}{\left | \vec{c} \right |} \right )$

$\vec{y}=(l\hat{a}+m\hat{b}+n\hat{c})$

Now projection of y along x is given as

$=\vec{y}.\dfrac{\vec{x}}{\left | \vec{x} \right |}$

$=\vec{y}.\hat{x}$

$=(l\hat{a}+m\hat{b}+n\hat{c}).\left ( \dfrac{\hat{a}}{\sqrt{2}}+\dfrac{\hat{b}}{\sqrt{2}} \right )$

$=\dfrac{l}{\sqrt{2}}+\dfrac{m}{\sqrt{2}}$

$=\dfrac{l+m}{\sqrt{2}}$

Ans (D)

$\dfrac{l+m}{\sqrt{2}}$

Using

$\vec{x}=\hat{a}+\hat{b}$

$\hat{x}=\dfrac{\hat{a+\hat{b}}}{\left | \hat{a}+\hat{b} \right |}=\dfrac{1}{\sqrt{2}}(\hat{a}+\hat{b})$

$\left | \hat{a}+\hat{b} \right |^{2}=\left | \hat{a} \right |^{2}+\left | \hat{b} \right |^{2}+2\hat{a}.\hat{b}$

$= 1+1+0=2$

Let

$\overline {a}, \overline {b}$ be two noncollinear vectors. If

$\overline {OA}=(x+4y)\overline {a}+(2x+y+1)\overline {b}, \overline {OB}=(y-2x+2)\overline {a}+(2x-3y-1)\overline {b}$ and

$3\overline {OA}=2\overline{OB}$ , then

$(x,y)=$

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$(1,2)$
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$(1,-2)$
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$(2,-1)$
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$(-2,-1)$

The projection of

$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ on the vector

$\vec{b}=\hat{i}+2\hat{j}-\hat{k}$ is?

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$\sqrt{\dfrac{2}{3}}$
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$\dfrac{2}{\sqrt{21}}$
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$\sqrt{\dfrac{3}{2}}$
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$\dfrac{5}{\sqrt{21}}$

Explanation

The given vectors are

$\vec { a } =\hat { i } +2\hat { j } +3\hat { k }$ and

$\vec { b } =\hat { i } +2\hat { j } -\hat { k }$

Let us first find the dot product of the given vectors as shown below:

$\vec { a } \cdot \vec { b } =(1\times 1)+(2\times 2)+(3\times -1)=1+4-3=5-3=2$

Now, we find the magnitude of the vector

$\vec { b }$ as follows:

$\left| \vec { b } \right| =\sqrt { { 1 }^{ 2 }+2^{ 2 }+(-1)^{ 2 } } =\sqrt { 1+4+1 } =\sqrt { 6 }$

We know that the projection of

$\vec { a }$ on

$\vec { b }$ is

$\dfrac { \vec { a } \cdot \vec { b } }{ \left| \vec { b } \right| }$ , therefore, we have:

$\dfrac { \vec { a } \cdot \vec { b } }{ \left| \vec { b } \right| } =\dfrac { 2 }{ \sqrt { 6 } } =\dfrac { 2 }{ \sqrt { 2\times 3 } } =\dfrac { 2 }{ \sqrt { 2 } \times \sqrt { 3 } } =\dfrac { \sqrt { 2 } }{ \sqrt { 3 } } =\sqrt { \dfrac { 2 }{ 3 } }$

Hence, the projection of the vector

$\vec { a }$ on

$\vec { b }$ is

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

Let

$\hat{a}, \hat{b}$ and

$\hat{c}$ be three unit vectors such that

$\hat{a}=\hat{b}+(\hat{b}\times \hat{c})$ , then the possible value(s) of

${|\hat{a}+\hat{b}+\hat{c}|}^{2}$ can be:

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$1$
0%
$4$
0%
$16$
0%
$9$

$D\vec A = \vec a,$

$A\vec B = \vec b$ and

$C\vec B = k\vec a$ where

$k < 0$ and X, Y are the mid-points of DB & DC respectively, such that

$\left| {\vec a} \right| = 17\& \left| {X\vec Y} \right| = 4$ , then k equal to -

0%
${8 \over {17}}$
0%
${{13} \over {17}}$
0%
${{25} \over {17}}$
0%
${4 \over {17}}$

$O$ is the origin in the Cartesian plane. From the origin

$O$ take point

$A$ in the North-east direction such that

$|\overline {OA}|=5,B$ is a point in the North-west direction such that

$|\overline {OB}|=5$ .
Then

$|\overline {OA}-\overline {OB}|$ is.

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$25$
0%
$5\sqrt2$
0%
$10\sqrt5$
0%
$\sqrt5$

The angles of a triangles whose two sides are represented by vectors

$\sqrt(\vec { a } \times \vec { b })$ and

$\vec{b}-(\vec { a } \vec { b })\vec{a}$ are in the ratio

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$1:2:3$
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$1:1:2$
0%
$1:3:5$
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$1:2:7$

The vectors

$\vec{u}=\begin{bmatrix} 6 \\ -3 \\ 2 \end{bmatrix}; \vec{v}=\begin{bmatrix} 2 \\ 6 \\ 3 \end{bmatrix}; \vec{w}=\begin{bmatrix} 3 \\ 2 \\ -6 \end{bmatrix}$

0%
form a left handed system
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form a right handed system
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are linearly independent
0%
are such that each is perpendicular to the plane containing the other two.

$\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular unit vectors and

$\bar{d}$ is a unit vector equally inclined to each other of

$\bar{a}, \bar{b}$ and

$\bar{c}$ at an angle of

$60^o$ . Then

$|\bar{a}+\bar{b}+\bar{c}+\bar{d}|^2=?$

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

${\textbf{Step -1: Solve the given equations}}{\textbf{.}}$

$\left| {\overline a } \right| = \left| {\overline b } \right| = \left| {\overline c } \right| = \left| {\overline d } \right| = 1$

${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + {\left| {\overrightarrow d } \right|^2} + 2\left( {\overrightarrow {a.} \overrightarrow b + \overrightarrow {a.} \overrightarrow {c.} + \overrightarrow {a.} \overrightarrow {d.} + \overrightarrow {b.} \overrightarrow {c.} + \overrightarrow {b.} \overrightarrow {a.} + \overrightarrow {c.} \overrightarrow {d.} } \right)$

$= 1 + 1 + 1 + 1 + 2\left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right]$

$+ \left[ {\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow b } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right]$

$= 4 + 2\left[ {0 + 0 + 1 \times \dfrac{1}{2} + 0 + 1 \times 1 \times \dfrac{1}{2} + 1 \times 1 \times \dfrac{1}{2}} \right]$

$= 4 + 2\left[ {\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}} \right]$

$= 4 + 2 \times \dfrac{3}{2}$

$= 4 + 3$

$= 7$

${\textbf{Hence, the value of }}{\mathbf{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right|^2} = 7.}$

Given unit vectors

$\hat {m}, \hat {n}$ and

$\hat {p}$ such that

$\left( \widehat { \hat { m } \hat { n } } \right) =\hat { p } \widehat { } \left( \hat { m } \times \hat { n } \right) =\alpha$ then the value of

$[\hat { n } \hat { p } \hat { m } ]$ in terms of

$\alpha$ is :

0%
$\sin \alpha$
0%
$\cos \alpha$
0%
$\sin \alpha \cos \alpha$
0%
$\sin^{2} \alpha$

$\left| {\overline x } \right| = \left| {\overline y } \right| = 1,\,\overline x \bot \overline y ,\,\left| {\overline x + \overline y } \right| =$

0%
$\sqrt 3$
0%
$\sqrt 2$
0%
$1$
0%
$0$

Which of the following is not essential for the three vectors to zero resultant?

0%
The resultant of any two vectors should be equal and opposite to the third vector
0%
They should lie in the same plane
0%
They should act the sides of a parallelogram
0%
It should be possible to represent them by the three sides of triangle taken in same order

The position vector of a point

$C$ with respect to

$B$ is

$\hat { i } + \hat { j }$ and that of B with respect to A is

$\hat { i } - \hat { j }$ . The position vector of

$C$ with respect to

$A$ is

0%
$\hat { 2i }$
0%
$-\hat { 2i }$
0%
$\hat { 2j }$
0%
$-\hat { 2j }$

$\vec{a}, \vec{b}, \vec{c}$ are unit vectors, then the value of

$|\vec{a}-2\vec{b}|^2+|\vec{b}-2\vec{c}|^2+|\vec{c}-2\vec{a}|^2$ does not exceed to?

0%
$9$
0%
$12$
0%
$18$
0%
$21$

D,E and F are the mid-points of the sides BC,CA and AB respectively of

$\Delta ABC$ and G is the centroid of the triangle, then

$\vec{GD}+\vec{GE}+\vec{GF}=$

0%
$\vec{0}$
0%
$2\vec{AB}$
0%
$2\vec{GA}$
0%
$2\vec{GC}$

The vector

$\bar { i } +x\bar { j } +3\bar { k }$ is rotated through an angle

$\theta$ and doubled in magnitude, then it becomes

$4\bar { i } +\left( 4x-2 \right) \bar { j } +2\bar { k }$ . The value of x is _________.

0%
$\left\{ -\frac { 2 }{ 3 } ,2 \right\}$
0%
$\left\{ \frac { 1 }{ 3 } ,2 \right\}$
0%
$\left\{ \frac { 2 }{ 3 } ,2 \right\}$
0%
$\left\{ 2,7 \right\}$

Forces

$3 \vec { OA }$ ,

$5 \vec { OB }$ act along OA and OB. If their resultant passes through C on AB, then :

0%
C is mid-point of AB
0%
C divides AB in the ratio $2:1$
0%
$3AC=5CB$
0%
$2AC=3CB$

Explanation

Draw

$ON$ perpendicular to

$AB$

Let

$\hat{i}$ be the unit vector along

$ON$

The resultant force

$\overrightarrow{R}=3\overrightarrow{A}+5\overrightarrow{B}$ ......

$\left(1\right)$

The angles between

$\hat{i}$ and the forces

$\overrightarrow{R}$ ,

$3\overrightarrow{A}$ and

$5\overrightarrow{B}$ are

$\angle{CON},\angle{AON}$ and

$\angle{BON}$ respectively.

$\overrightarrow{R}.\hat{i}=3\overrightarrow{A}.\hat{i}+5\overrightarrow{B}.\hat{i}$

$\overrightarrow{R}.1.\cos{\angle{CON}}=3.\overrightarrow{OA}.1.\cos{\angle{AON}}+5\overrightarrow{OB}.1.\cos{\angle{BON}}$

$\overrightarrow{R}.\dfrac{ON}{OC}=3\overrightarrow{OA}\times \dfrac{ON}{OA}+5\overrightarrow{OB}\times \dfrac{ON}{OB}$

$\Rightarrow \dfrac{\overrightarrow{R}}{OC}=3+5$

$\Rightarrow \overrightarrow{R}=8\overrightarrow{OC}$

We know that

$\overrightarrow{OA}=\overrightarrow{OC}+\overrightarrow{CA}$

$\Rightarrow 3\overrightarrow{OA}=3\overrightarrow{OC}+3\overrightarrow{CA}$ .......

$\left(1\right)$ by multiplying the complete equation by

$3$

We know that

$\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}$

$\Rightarrow 5\overrightarrow{OB}=5\overrightarrow{OC}+5\overrightarrow{CB}$ .......

$\left(2\right)$ by multiplying the complete equation by

$5$

Eqn

$\left(1\right)+\left(2\right)$

$\Rightarrow 3\overrightarrow{OA}+5\overrightarrow{OB}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$

$\Rightarrow \overrightarrow{R}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$

$\Rightarrow 8\overrightarrow{OC}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$

$\Rightarrow \left|3\overrightarrow{CA}\right|=\left|5\overrightarrow{CB}\right|$

$\therefore 3AC=5CB$

$\vec a = 4 i + 5 j - k , \vec { b } = i - 4 j + 5 k , \vec c = 3 i + j - k$ such that

$\vec { p } \perp \vec { a } , \vec { p } \perp \overline { b }$ and

$\vec { p . c } = 21 ,$ then

$p =$

0%
$6$ ( i + j + k )$$
0%
$7$ ( i + j + k )$$
0%
$6$ ( i - j - k )$$
0%
$7$ ( i - j - k )$$

The position vector of two points A and B are 6a+2b and a-3b. If a point C divides AB in the ratio 3 : 2, then the position vector of C is

0%
3a-b
0%
3a+b
0%
a+b
0%
a-b

$\vec { a } =2\hat { I } +\hat { k } ,\vec { b } ={ b }_{ 1 }\hat { I } +{ b }_{ 2 }\hat { J+ } { b }_{ 3 }\hat { k, } \vec { a } \times \vec { b } =5\hat { I } +2\hat { J } -12\hat { k, } \hat { a } \hat { b } =11\quad then\quad { b }_{ 1 }+{ b }_{ 2 }+{ b }_{ 3 }=$

0%
3
0%
5
0%
7
0%
9

The X & Y Component of vector A have numerical values 6 each & that of ( A +B) have numerical values 10 and 9 .What is the numerical value of B ?

0%
2
0%
3
0%
4
0%
5

$a,b$ and

$c$ are position vector of

$A,B$ and

$C$ respectively of

$\triangle ABC$ and if

$|a-b|=4,|b-c|=2, |c-a|=3$ , then the distance between the centroid and incentre of

$\triangle ABC$ is

0%
$1$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$

Vector equation of the plane

$\vec{r}=\hat{i}-\hat{j}+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}+2\hat{j}+3\hat{k})$ in the scalar dot product from is

0%
$\vec{r}.(5\hat{i}-2\hat{j}+3\hat{k})=7$
0%
$\vec{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$
0%
$\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=7$
0%
$\vec{r}.(5\hat{i}+2\hat{j}+3\hat{k})=7$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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