MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 7

$2\overrightarrow { a }\cdot \overrightarrow { b } =\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right|$ then the angle between

$\vec { a }$ and

$\vec { b }$ is

0%
${30}^{o}$
0%
${0}^{o}$
0%
${90}^{o}$
0%
${60}^{o}$

Explanation

$2\overrightarrow { a }\cdot \overrightarrow { b } = \left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right|$

But we know that

$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos\theta$

$\therefore \cos\theta=\dfrac{1}{2}$

$\Rightarrow \theta=60^o$

Here

$\theta$ is the angle between both the vectors

For non-zero vectors

$\overrightarrow{a}$ and

$\overrightarrow{b}$ if

$|\overrightarrow{a}+\overrightarrow{b}| < |\overrightarrow{a}-\overrightarrow{b}|$ , then

$\overrightarrow{a}$ and

$\overrightarrow{b}$ are

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Collinear
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Perpendicular to each other
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Inclined at an acute angle
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Inclined at an obtuse angle

Explanation

$|\vec{a}+\vec{b}|<|\vec{a}-\vec{b}|$

$a^{2}+b^{2}+2ab\cos\theta<a^{2}+b^{2}-2ab\cos\theta$ where

$\theta$ is the included angle between the two vectors.
Hence

$4ab\cos\theta<0$
Since

$a,b>0$ (magnitude of a vector is always a positive quantity).
Hence

$\cos\theta<0$
Or

$\theta\in \left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)$ . Hence

$\theta$ is an obtuse angle.

Let

$\vec {a} = i + 2j + k, \vec {b} = i - j + k$ and

$\vec {c} = i + j - k$ , a vector in the plane

$\vec {a}$ and

$\vec {b}$ whose projection on

$\vec {c}$ is

$\dfrac {1}{\sqrt {3}}$ is _____

0%
$3i + j - 3k$
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$4i + j - 4k$
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$i + j - 2k$
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$2i + j + 2k$

Explanation

A vector in the plane of

$\vec{a}$ and

$\vec{b}$ would be

$m\vec{a} + n\vec{b}$

i.e.

$(m + n)i + (2m - n)j + (m + n)k$

Its projection on

$\vec{c}$ will be

$\cfrac{(m\vec{a} + n\vec{b}).\vec{c}}{|c|}$

$\therefore \cfrac{1}{\sqrt{3}} = \cfrac{[(m + n)i + (2m - n)j + (m + n)k].(i + j - k)}{\sqrt{3}}$

$\Rightarrow 1 = m + n + 2m - n - m - n$

$\Rightarrow 1 = 2m - n$

This relation is satisfied by

$m = 1, n = 1$

Thus, the required vector can be

$2i + j + 2k$

The value of

$x$ if

$x(\hat { i } +\hat { j } +\hat { k } )$ is a unit vector is

0%
$\pm \cfrac { 1 }{ \sqrt { 3 } }$
0%
$\pm \sqrt { 3 }$
0%
$\pm 3$
0%
$\pm \cfrac{1}{3}$

Explanation

$x(\hat i+\hat j+\hat k)$ is an unit vector then

$|x(\hat i+\hat j+\hat k)|=1$

$\Rightarrow \bigg|x\sqrt{1^2+1^2+1^2} \bigg|=1$

$\Rightarrow \sqrt 3|x|=1$

$\Rightarrow |x|=\dfrac{1}{\sqrt 3}$

$\therefore x=\pm \dfrac{1}{\sqrt 3}$

$\vec { a }$ and

$\vec { b }$ are two unit vector and

$\theta$ is the angle between them, then

$\left( \vec { a } +\vec { b } \right)$ is a unit vector if

$\theta =$

0%
$\dfrac { \pi }{ 3 }$
0%
$\dfrac { \pi }{ 4 }$
0%
$\dfrac { \pi }{ 2 }$
0%
$\dfrac { 2\pi }{ 3 }$

Explanation

Given

$|\vec a+\vec b|=1$

$\Rightarrow |\vec a|^2+2\vec a\cdot \vec b+|\vec b|^2=1$ , square both sides

$\Rightarrow 1+2|\vec a||\vec {b}|\cos \theta+1=1$

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

$\therefore \theta=\dfrac{2\pi}{3}$

The position vector of a point which is

$2$ units away from

$3\hat i + 4\hat j + 2\hat k$ along the

$x$ -axis is:

0%
$\hat i + 4\hat j + 2\hat k$
0%
$3\hat i + 2\hat j + 2\hat k$
0%
$5\hat i + 4\hat j + 2\hat k$
0%
$3\hat i + 4\hat j + \hat k$

Explanation

A vector that starts from the origin (O) is called a position vector.

Consider the above image:

in the given equation

$3\hat i+4\hat j+2\hat k$ ......(1)

we have to take away 2 units along x axis it may be left of thea xis or right of the axis , hence this question have 2 answers:

$\hat i+4\hat j+2\hat k$

$\Rightarrow$ When 2units (left) away form eqn (1)

$5\hat i+4\hat j+2\hat k$

$\Rightarrow$ When 2units (right) away form eqn (1).

The projection of

$\vec {i} - \vec {j}$ on

$z$ -axis is

0%
$0$
0%
$1$
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$-1$
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$2$

Explanation

The projection of

$\vec{a}=i-j$ on

$\vec{b}=k$ is given by

$a_{1}=\dfrac{\vec{a}.\vec{b}}{|\vec{b}|}$

$=\dfrac{(i-j).(k)}{1}$

$=0$

The vector

$b = 3j + 4k$ is to be written as the sum of a vector

$b_{1}$ parallel to

$a = i + j$ and a vector

$b_{2}$ perpendicular to

$a$ . Then

$b_{1}$ is equal to

0%
$\dfrac {3}{2}(i + j)$
0%
$\dfrac {2}{3}(i + j)$
0%
$\dfrac {1}{2}(i + j)$
0%
$\dfrac {1}{3}(i + j)$

Explanation

$b_{1} \parallel a\Rightarrow b_{1} = a(i + j)$

$b_{2} = b - b_{1} = (3 - a) i - aj + 4k$
Also,

$b_{3} - a = 0$

$\Rightarrow (3 - a) - a=0\Rightarrow a = \dfrac {3}{2}$

$\therefore b_{1} = \dfrac {3}{2}(i + j)$

$\vec a, \vec b$ and

$\vec c$ are three non-coplanar vectors, then

$(\vec a +\vec b -\vec c) \cdot [(\vec a - \vec b) \times (\vec b - \vec c)$ equals

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$\vec 0$
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$\vec a.(\vec b\times \vec c)$
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$\vec a.(\vec c\times \vec b)$
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$3\vec a.(\vec b\times \vec c)$

Explanation

$[a + b - c] \cdot [(a - b)\times (b - c)$

$= (a + b - c)\cdot [a\times b - a\times c - b\times b + b\times c]$

Since,

$(b \times b=0)$

$\therefore [a + b - c] \cdot [(a - b)\times (b - c)]$

$= a.(a\times b) - a.(a\times c) + a.(b\times c) + b.(a\times b) - b(a\times c) + b.(b\times c) - c.(a\times b) + c.(a\times c) - c.(b\times c)$

$= a.(b\times c) - b.(a\times c) - c.(a\times b)$ ........ (since, a,b and c are non-

coplanar)

$=[a\ b\ c] - [b\ a\ c] - [c\ a\ b]$

$=[a\ b\ c] + [a\ b\ c] - [a\ b\ c]$

$=[a\ b\ c] = a.(b\times c)$

Hence, option B is correct.

$\overrightarrow { a }$ is vector of magnitude

$x$ ,

$m$ is non-zero scalar and

$m\overrightarrow { a }$ is a unit vector then x in terms of m is:

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$m=x$
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$x=\left| m \right|$
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$x=\cfrac { 1 }{ \left| m \right| }$
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$x=m/2$

Explanation

Given

$|m\vec{a}|=1$

$\Rightarrow |m||\vec{a}|=1$

$\Rightarrow |m|x=1$

$\Rightarrow x=\dfrac{1}{|m|}$

Remember: modulus can never be negative

$\vec { a } ,\vec { b } ,\vec { c }$ are mutually perpendicular unit vectors, then

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

0%
$3$
0%
$\sqrt { 3 }$
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Zero
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$1$

Explanation

Solution:

Given that:

$\vec a,\vec b,\vec c$ are mutually perpendicular .i.e

$\vec a.\vec b=0,\vec b.\vec c=0$ and

$\vec c.\vec a=0$

Also,

$|\vec a|=|\vec b|=|\vec c|=1$

Solution:

$|\vec a+\vec b+\vec c|=\sqrt{|\vec a|^2+|\vec b|^2+|\vec c|^2+2\vec a.\vec b+2\vec b.\vec c+2\vec c.\vec a}$

$=\sqrt{1+1+1+0+0+0}$

$=\sqrt{3}$

Hence, B is the correct option.

$\overrightarrow{b}$ and

$\overrightarrow{c}$ are the position vectors of the points B and C respectively, then the position vector of the point D such that

$\overrightarrow{BD}=4\overrightarrow{BC}$ is

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$4(\overrightarrow{c}-\overrightarrow{b})$
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$-4(\overrightarrow{c}-\overrightarrow{b})$
0%
$4\overrightarrow{c}-3\overrightarrow{b}$
0%
$4\overrightarrow{c}+3\overrightarrow{b}$

Explanation

Solution:

Given that:

$\overrightarrow{BD}=4\overrightarrow{BC}$

It means

$D$ divides the join of

$BC$ externally in the ratio

$4:3.$

$\therefore$ Position vector of

$D=\cfrac{4\overrightarrow c-3\overrightarrow b}{4-3}=4\overrightarrow c-3\overrightarrow b$

Hence, C is the correct option.

Consider the vectors

$\bar{a}=\hat{i}-2\hat{j}+\hat{k}$ and

$\bar{b}=4\hat{i}-4\hat{j}+7\hat{k}$

Find the scalar projection of

$\bar{a}$ on

$\bar{b}.$

0%
1
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$\dfrac{19}{9}$
0%
$\dfrac{17}{9}$
0%
$\dfrac{23}{9}$

Explanation

$\bar{a}=\hat{i}-2\hat{j}+\hat{k}$ and

$\bar{b}=4\hat{i}-4\hat{j}+7\hat{k}$

Scalar projection of

$a$ on

$b$

$P=\cfrac { \vec { a } .\vec { b } }{ \left| \vec { b } \right| } \\ P=\cfrac { (\hat { i } -2\hat { j } +\hat { k } ).(4\hat { i } -4\hat { j } +7\hat { k } ) }{ \sqrt { { 4 }^{ 2 }+{ (-4) }^{ 2 }+{ 7 }^{ 2 } } } \\ \hat { i } .\hat { i } =\hat { j } .\hat { j } =\hat { k } .\hat { k } =1\\ P=\cfrac { 4+8+7 }{ 9 } \\ P=\cfrac { 19 }{ 9 }$

The magnitude of the scalar

$p$ for which the vector

$p\left( -3\hat { i } -2\hat { j } +13\hat { k } \right)$ is of unit length is:

0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{64}$
0%
$\sqrt { 182 }$
0%
$\dfrac{1}{\sqrt { 182 }}$

Explanation

For given vector to be unit vector its magnitude must be equal to one.
Now modulus of given vector

$=p\sqrt { { (-3) }^{ 2 }+{ (-2) }^{ 2 }+{ 13 }^{ 2 } } =p\sqrt { 182 } =1\\ \Rightarrow p=\dfrac { 1 }{ \sqrt { 182 } }$
Option D is correct.

Let

$\vec{v_1}, \vec{v_2}, \vec{v_3} , \vec{v_4}$ be unit vectors in the xy - plane, one each in the interior of the four quadrants. Which of the following statements is necesserily true?

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$\vec{v_1}, \vec{v_2}, \vec{v_3} , \vec{v_4} = 0$
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There exist i, j with $1 \le i \le j \le 4$ such that $\vec{v_i} + \vec{v_j}$ is in the first quadrant.
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There exist i, j with $1 \le i \le j \le 4$ such that $\vec{v_i} + \vec{v_j} < 0$
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There exist i, j with $1 \le i \le j \le 4$ such that $\vec{v_i} + \vec{v_j} >0$

Explanation

A zero vector can never be a unit vector

$\therefore$ option $A$ is not correct

Let as assume four unit vectors in the $xy$ plane as shown in the figure

$\underset { { V }_{ 1 } }{ \rightarrow } =\dfrac { i }{ \sqrt { 2 } } +\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 2 } }{ \rightarrow } =-\dfrac { i }{ \sqrt { 2 } } +\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 3 } }{ \rightarrow } =-\dfrac { i }{ \sqrt { 2 } } -\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 4 } }{ \rightarrow } =\dfrac { i }{ \sqrt { 2 } } -\dfrac { j }{ \sqrt { 2 } }$

If we add up any of the two vectors it will either lies on the $x$ aisx or the $y$ axis, none of them lies in the first quardrant

$\therefore$ option $B$ is not correct

Sum of any two vectors less than or greater than zero does not make any sense

$\therefore$ options $C$ and $D$ are incorrect

So none of the options are correct.

If the position vector

$\overrightarrow{a}$ of the point

$(5, n)$ is such that

$|\overrightarrow{a}|=13$ , then the value/values of n be

0%
$\pm 8$
0%
$\pm 12$
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8 only
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12 only

Explanation

Solution:

Given that:

$|\overrightarrow a|=13$

We have,

$\overrightarrow a=5\hat i+n\hat j$

or,

$|\overrightarrow a|=\sqrt{5^2+n^2}$

or,

$13=\sqrt{25+n^2}$

or,

$n^2=169-25=144$

or,

$n=\pm12$

Hence, B is the correct option.

$\overrightarrow a$ and

$\overrightarrow b$ are unit vectors, then angle between

$\overrightarrow a$ and

$\overrightarrow b$ for

$\sqrt 3 \overrightarrow a - \overrightarrow b$ to be unit vector is

0%
$60^o$
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$45^o$
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$30^o$
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$90^o$

Explanation

${\textbf{Step-1: Apply unit vector property.}}$

${\text{Given,}}$

$\bar a$

${\text{and}}$

$\bar b$

${\text{are unit vectors and}}$

$( \sqrt 3 \bar a - \bar b)$

${\text{is also unit vector.}}$

${\text{Let angle be}}$

$\theta$

$\Rightarrow \bar a. \bar b =| \bar a| | \bar b| cos \theta$

${\text{.....Dot product of two vectors}}$

${\text{But,}}$

$\bar a$

${\text{and}}$

$\bar b$

${\text{are unit vectors.}}$

$|\bar a |=$

$| \bar b |= 1$

$\Rightarrow \bar a. \bar b =cos \theta$

$....(1)$

$( \sqrt 3 \bar a - \bar b) =1$

${\text{also unit vector.}}$

${\textbf{Step-2: Squaring on both side and find angle.}}$

$( \sqrt 3 \bar a - \bar b)^2 =1$

$.......\because (a-b)^2 = a^2-2ab +b^2$

$\Rightarrow 3(1) + (1) - 2 \sqrt3 cos \theta = 1$

$....(1)$

$\Rightarrow 4 - 2 \sqrt3 cos \theta = 1$

$\Rightarrow 2 \sqrt3 cos \theta = 3$

$\Rightarrow cos \theta = \dfrac{\sqrt3}{2} = cos 30 ^\circ$

$\Rightarrow \theta = 30 ^\circ$

${\textbf{Hence, the option (C) is correct.}}$

$a\times b=c$ ,

$b\times c=a$ and

$a,b,c$ be the mod of the vectors

$a,b,c$ respectively, then

0%
$a=1, b=1$
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$c=1, a=1$
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$a\cdot \left( b\times c \right) =1$
0%
$b=1,c=a$

Explanation

Since,

$a\times b=c$
Thus

$\left( b\times c \right) \times b=c$

$\Rightarrow \left( b\cdot b \right) c-\left( b\cdot c \right) b=c$

$\Rightarrow \left( b\cdot b \right) =1$ ,

$c\cdot b=0$

$\Rightarrow { b }^{ 2 }=1$ i.e.,

$b=1$

$b$ is a unit vector
Therefore,

$\left| c \right| =\left| a \right| \Rightarrow c=a$

Let

$u, v$ and

$w$ be such that

$\left| u \right| =1, \left| v \right| =3$ and

$\left| w \right| =2$ . If the projection of

$v$ along

$u$ is equal to that of

$w$ along

$u$ and vectors

$v$ and

$w$ are perpendicular to each other, then

$\left| u-v+w \right|$ equals

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

Explanation

Given,

$v\cdot u=w\cdot u$ and

$v \perp w$

$\Rightarrow v\cdot w=0$
Now, consider

${ \left| u-v+w \right| }^{ 2 }={ \left| u \right| }^{ 2 }+{ \left| v \right| }^{ 2 }+{ \left| w \right| }^{ 2 }-2u\cdot v-2w\cdot v+2u\cdot w$

$=1+9+4=14$

$\Rightarrow \left| u-v+w \right| =\sqrt { 14 }$

A vector

$R$ is given by

$R=A\times \left( B\times C \right)$ , which of the following is true?

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$R$ must be perpendicular to $B$
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$R$ is parallel to $A$
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$R$ must be parallel to $B$
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None of the above

Explanation

Given that:

$R=A\times(B\times C)$

Geometrical representation of the above is

$R$ must be perpendicular to

$A$ and also

$R$ must be perpendicular to

$(B\times C).$

Therefore, D is the correct option.

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

$r\times a=b\times a, r\times b=a\times b$ , then a unit vector in the direction of

${r}$ is?

0%
$\displaystyle\frac{1}{\sqrt{11}}(\hat{i}+3\hat{j}-\hat{k})$
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$\displaystyle \frac{1}{\sqrt{11}}(\hat{i}-3\hat{j}+\hat{k})$
0%
$\displaystyle \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
0%
None of these

Explanation

We have,

$r\times a =b\times a$ and

$r\times b=a\times b$

$\Rightarrow r\times a =-(r\times b)$

$\Rightarrow r\times (a+b)=0$

$\Rightarrow$ r is parallel to

$a+b$

$\Rightarrow r=\lambda(a+b)$

$\Rightarrow r=\lambda(\hat{i}+3\hat{j}-\hat{k})$

$\Rightarrow |r|=\sqrt{11}\lambda$

$\therefore$ Required vector

$=\displaystyle\frac{r}{|r|}=\frac{1}{\sqrt{11}}(\hat{i}+3\hat{j}-\hat{k})$ .

The resultant of

$P$ and

$Q$ is

$R$ . If

$Q$ is doubled,

$R$ is also doubled and if

$Q$ is reversed,

$R$ is again doubled. Then,

$P^{2} : Q^{2} : R^{2}$ given by

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

Explanation

$R^{2} = P^{2} + Q^{2} = 2PQ\cos \theta$

$4R^{2} = P^{2} + 4Q^{2} + 4PQ \cos \theta$

$4R^{2} = P^{2} + 4Q^{2} - 2PQ \cos \theta$
On solving,

$\dfrac {P^{2}}{-6} = \dfrac {Q^{2}}{-9} = \dfrac {R^{2}}{-6}$

$\Rightarrow P^{2} : Q^{2} : R^{2}$ as

$2 : 3 : 2$ .

If the scalar projection of the vector

$xi-j+k$ on the vector

$2i-j+5k$ is

$\dfrac { 1 }{ \sqrt { 30 } }$ then value of

$x$ is equal to

0%
$\dfrac { -5 }{ 2 }$
0%
$6$
0%
$-6$
0%
$3$

Explanation

Scalar projection of $\vec { a }$ on $\vec { b }$ $=\dfrac { \left| \vec { a } .\vec { b } \right| }{ \left| \vec { b } \right| }$

let $\vec { a } \ =xi-j+k$ and $\vec { b } =2i-j+5k$

Then projection of $\vec { a }$ on $\vec { b }$ $=\dfrac { (xi-j+k).(2i-j+5k) }{ \left| 2i-j+5k \right| }$

Given scalar projection $=\dfrac { 1 }{ \sqrt { 30 } }$

$\Rightarrow \dfrac { 1 }{ \sqrt { 30 } } =\dfrac { 2x+1+5 }{ \sqrt { { 2 }^{ 2 }+{ (-1) }^{ 2 }+{ 5 }^{ 2 } } } \\ \Rightarrow \dfrac { 1 }{ \sqrt { 30 } } =\dfrac { 2x+1+5 }{ \sqrt { 30 } } \\ \Rightarrow 2x+1+5=1\\ \Rightarrow x=-\dfrac { 5 }{ 2 }$

In the given diagram, if PQ

$=$ A, QR

$=$ B and RS

$=$ C, then PS will be equal to :

0%
$A-B+C$
0%
$A+B-C$
0%
$A+B+C$
0%
$A-B-C$
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$-A-B-C$

Explanation

According to polygon law of vector addition

$A+B=PR$
and

$PR+RS=PS$

$\Rightarrow A+B+C=PS$ .

Let

$\square PQRS$ be a quadrilateral. If

$M$ and

$N$ are midpoints of the sides

$PQ$ and

$RS$ respectively then

$\overline {PS} + \overline {QR} =$

0%
$3\overline {MN}$
0%
$4\overline {MN}$
0%
$2\overline {MN}$
0%
$2\overline {NM}$

Explanation

Let

$\overline{m}, \overline{n}, \overline{p}, \overline{q}, \overline{r}$ and

$\overline{s}$ be position vectors of

$M,N,P,Q,R$ and

$S$ respectively.

$M$ and

$N$ are midpoints of

$PQ$ and

$RS$

$\therefore \overline {m} = \dfrac {\overline {p} + \overline {q}}{2}$ and

$\overline {n} = \dfrac {\overline {r} + \overline {s}}{2}$

$\overline {PS} + \overline {QR} = \overline {s} - \overline {p} + \overline {r} - \overline {q}$

$= (\overline {r} + \overline {s}) - (\overline {p} + \overline {q})$

$= 2\overline {n} - 2\overline {m}$

$= 2(\overline {n} - \overline {m})$

$= 2\overline {MN}$

$\vec{a}$ and

$\vec{b}$ are the vectors determined by two adjacent sides of regular hexagon, then vector

$EF$ is

0%
$(\vec{a} + \vec{b})$
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$(\vec{a} - \vec{b})$
0%
$2 \vec{a}$
0%
$2\vec{b}$

Explanation

FABCDE is a regular hexgon.

Let

$\overline{FA}=\vec{a}$ and

$\overline{AB}=\vec{b}$ , join

$FB$ and

$FC$ , we have

$\overline{FB}=\overline{FA}+\overline{AB}=\vec{a}+\vec{b}$

Since,

$\overline{FC}$ is parallel to

$\overline{AB}$ and double of

$\overline{AB}$

$\therefore \overline{FC}=2\overline{AB}=2\vec{b}$

Now,

$\overline{BC}=FC-FB$

$=2\vec{b}-(\vec{a}+\vec{b})=\vec{b}-\vec{a}$

$\overline{CD}=-\overline{FA}=-\vec{a}$ and

$\overline{DE}=-\overline{AB}=-\vec{b}$

Also,

$\overline{EF}=-\overline{BC}=-(\vec{b}-\vec{a})=\vec{a}-\vec{b}$

$p=\hat {i} + \hat{j}, q=4\hat{k}-\hat{j}$ and

$r=\hat{i}+\hat{k}$ , then the unit vector in the direction of

$3p+q-2r$ is

0%
$\dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$
0%
$\dfrac{1}{3}(\hat{i}-2\hat{j}-2\hat{k})$
0%
$\dfrac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})$
0%
$\hat{i}-2\hat{j}+2\hat{k}$

Explanation

$3p+q-2r=3(\hat{i}+\hat{j})+(4\hat{k}-\hat{j})-2(\hat{i}+\hat{k})$

$=\hat{i}+2\hat{j}+2\hat{k}$

$\therefore$ Unit vector in the direction of

$3p+q-2r = \dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$

$a.b=0$ and

$a+b$ makes an angle of

${ 60 }^{ o }$ with

$b$ , then

$\left| a \right|$ is equal to

0%
$0$
0%
$\cfrac { 1 }{ \sqrt { 3 } } \left| b \right|$
0%
$\cfrac { 1 }{ \left| b \right| }$
0%
$\left| b \right|$
0%
$\sqrt { 3 } \left| b \right|$

Explanation

We have

$a.b=0\Rightarrow a\bot b$
so, vectors

$a,b$ and

$a+b$ form a right angled triangle
In

$\triangle PQR\quad$

$\tan { { 60 }^{ o } } =\cfrac { \left| a \right| }{ \left| b \right| }$

$\Rightarrow \sqrt { 3 } =\cfrac { \left| a \right| }{ \left| b \right| }$

$\Rightarrow \left| a \right| =\sqrt { 3 } \left| b \right|$

$a\cdot b =0$ and

$a + b$ makes an angle

$60^o$ with

$a$ , then

0%
$|a| = 2 |b|$
0%
$2 |a| = |b|$
0%
$|a| = \sqrt 3 |b|$
0%
$|a| = |b|$
0%
$\sqrt 3 |a | = |b|$

Explanation

Given,

$a\cdot b = 0$
Now,

$(a + b)\cdot a = |a + b||a| \cos 60^o$

$\Rightarrow a\cdot a + b\cdot a = |a + b| |a| \dfrac{1}{2}$

$\Rightarrow |a|^2 + 0 = \dfrac{|a + b| |a|}{2}$

$\Rightarrow 2 |a| = |a + b|$
On squaring both sides, we get

$4 |a|^2 = |a|^2 + |b|^2 + 2 a\cdot b \cos \theta$

$4 |a|^2 = |a|^2 + |b|^2 + 2 |a| |b| \cos \theta$

$\Rightarrow 3 |a|^2 = |b|^2 + 0 (\because |a| |b| \cos \theta = a \cdot b = 0 )$

$\Rightarrow \sqrt 3 |a| = |b|$

Let

$ABCD$ be a parallelogram. If

$AB=\hat { i } +3\hat { j } +7\hat { k } , AD=2\hat { i } +3\hat { j } -5\hat { k }$ and

$p$ is a unit vector parallel to

$AC$ , then

$p$ is equal to

0%
$\dfrac { 1 }{ 3 } \left( 2\hat { i } +\hat { j } +2\hat { k } \right)$
0%
$\dfrac { 1 }{ 3 } \left( 2\hat { i } +2\hat { j } +2\hat { k } \right)$
0%
$\dfrac { 1 }{ 7 } \left( 3\hat { i } +6\hat { j } +2\hat { k } \right)$
0%
$\dfrac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } +3\hat { k } \right)$
0%
$\dfrac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right)$

Explanation

Given,

$AB=\hat i+3\hat j+7\hat k, AD=2\hat i+3\hat j-5\hat k$

$AC=AB+AD$

$=\left( \hat { i } +3\hat { j } +7\hat { k } \right) +\left( 2\hat { i } +3\hat { j } -5\hat { k } \right)$

$=3\hat { i } +6\hat { j } +2\hat { k }$
Now,

$p$ is a unit vector parallel to

$AC$ .
Then,

$p=\dfrac { AC }{ \left| AC \right| }$

$=\dfrac { 3\hat { i } +6\hat { j } +2\hat { k } }{ \sqrt { 9+36+4 } }$

$\Rightarrow p=\dfrac { 1 }{ 7 } \left( 3\hat { i } +6\hat { j } +2\hat { k } \right)$

Let

$u, v$ and

$w$ be vectors such that

$u + v + w = 0$ . If

$|u| = 3, |v| = 4$ and

$|w| = 5$ , then

$u . v + v . w + w.u$ is equal to

0%
$0$
0%
$-25$
0%
$25$
0%
$50$
0%
$47$

Explanation

Given,

$|u| = 3, |v| = 4$ and

$|w| = 5$

Also,

$u + v + w = 0$

On squaring both sides, we get

$|u|^2 + |v|^2 + |w|^2 + 2(u. v + v.w + w.u) = 0$

$\Rightarrow 3^2 + 4^2 + 5^2 + 2 (u.v + v.w + w.u) = 0$

$\Rightarrow 9 + 16 + 25 + 2(u.v + v.w + w.u) = 0$

$\Rightarrow u.v + v.w + w.u = - \dfrac{50}{2} = - 25$

Let

$P\left( 1,2,3 \right)$ and

$Q\left( -1,-2,-3 \right)$ be the two points and let

$O$ be the origin. Then,

$\left| PQ+OP \right|$ is equal to

0%
$\sqrt { 13 }$
0%
$\sqrt { 14 }$
0%
$\sqrt { 24 }$
0%
$\sqrt { 12 }$
0%
$\sqrt { 8 }$

Explanation

Given that,

$P=\left( 1,2,3 \right) , Q=\left( -1,-2,-3 \right)$ and

$O=\left( 0,0,0 \right)$ .

Then,

$PQ=\left( -1-1 \right) \hat { i } +\left( -2-2 \right) \hat { j } +\left( -3-3 \right) \hat { k }$

$=-2\hat { i } -4\hat { j } -6\hat { k }$
and

$OP=\left( 1-0 \right) \hat { i } +\left( 2-0 \right) \hat { j } +\left( 3-0 \right) \hat { k }$

$=\hat { i } +2\hat { j } +3\hat { k }$
Now,

$PQ+OP=\left( -2\hat { i } -4\hat { j } -6\hat { k } \right) +\left( \hat { i } +2\hat { j } +3\hat { k } \right)$

$=-\hat { i } -2\hat { j } -3\hat { k }$

$\Rightarrow \left| PQ+OP \right| =\sqrt { { \left( -1 \right) }^{ 2 }+{ \left( -2 \right) }^{ 2 }+{ \left( -3 \right) }^{ 2 } } =\sqrt { 14 }$

Alternate Method

$PQ=OQ-OP$

$\Rightarrow PQ+OP=OQ$
Now,

$OQ=-\hat { i } -2\hat { j } -3\hat { k }$

$\left| OQ \right| =\sqrt { { \left( -1 \right) }^{ 2 }+{ \left( -2 \right) }^{ 2 }+{ \left( -3 \right) }^{ 2 } }$

$=\sqrt { 14 }$
Therefore,

$\left| PQ+OP \right| =\left| OQ \right|$

$=\sqrt { 14 }$

$\widehat i + \widehat j, \widehat j + \widehat k, \widehat i + \widehat k$ are the position vectors of the vertices of a

$\Delta ABC$ taken in order, then

$\angle A$ is equal to

0%
$\dfrac {\pi}{2}$
0%
$\dfrac {\pi}{5}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{3}$

Explanation

Let position vector of the vertices are

$OA = \widehat i + \widehat j, OB = \widehat j + \widehat k$
and

$OC = \widehat i + \widehat k$
Now,

$AB = - \widehat i + \widehat k$
and

$AC = \widehat k - \widehat j$

$\therefore \cos \theta = \dfrac{(AB).(AC)}{|AC||AC|}$

$= \dfrac{(-\widehat i + \widehat k) . (\widehat k - \widehat j)}{\sqrt{1^2 + 1^2} \sqrt{1^2 + 1^2}}$

$= \dfrac{(\widehat k)^2}{\sqrt 2 \sqrt 2} = \dfrac{1}{2}$

$\Rightarrow \theta = \dfrac{\pi}{3}$

The angle between the two vectors

$\hat { i } +\hat { j } +\hat { k }$ and

$2\hat { i } -2\hat { j } +2\hat { k }$ is equal to

0%
$\cos ^{ -1 }{ \left( \dfrac { 2 }{ 3 } \right) }$
0%
$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 6 } \right) }$
0%
$\cos ^{ -1 }{ \left( \dfrac { 5 }{ 6 } \right) }$
0%
$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 18 } \right) }$
0%
$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) }$

Explanation

$\textbf{Step 1: Angle between two vectors }(\theta)$

Dot product between the two vectors is defined as:

$\vec{a}.\vec{b}= |\vec{a}||\vec{b}|\cos\theta$

$\Rightarrow \cos \theta = \dfrac {\vec {a} \cdot \vec {b}}{|\vec {a}| |\vec {b}|} = \dfrac {(\hat {i} + \hat {j} + \hat {k})\cdot (2\hat {i} - 2\hat {j} + 2\hat {k})}{\sqrt {3}\times \sqrt {4 + 4 + 4}}$

$= \dfrac {2 - 2 + 2}{\sqrt {3} \times \sqrt {12}} = \dfrac {1}{3}$

$\Rightarrow\theta = \cos^{-1} \left (\dfrac {1}{3}\right )$

Hence angle between vectors is

$\cos^{-1}\left (\dfrac {1}{3}\right )$

Option (E) correct.

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

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

$c=\hat { i } +\alpha \hat { j } +\beta \hat { k }$ are coplanar and

$\left| c \right| =\sqrt { 3 }$ , then

0%
$\alpha =\sqrt { 2 } ,\beta =1$
0%
$\alpha =1,\beta =\pm 1$
0%
$\alpha =\pm 1,\beta =1$
0%
$\alpha =\pm 1,\beta =-1$
0%
$\alpha =-1,\beta =\pm 1$

Explanation

Given that,

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

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

$c=\hat { i } +\alpha \hat { j } +\beta \hat { k }$
Since,

$a, b$ and

$c$ are coplanar.
Then,

$\begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix}=0$

$\Rightarrow 1\left( 3\beta -4\alpha \right) -1\left( 4\beta -4 \right) +1\left( 4\alpha -3 \right) =0$

$\Rightarrow 3\beta -4\alpha -4\beta +4+4\alpha -3=0$

$\Rightarrow \beta =1$
Also,

$\left| c \right| =\sqrt { 3 }$

$\Rightarrow \sqrt { 1+{ \alpha }^{ 2 }+{ \beta }^{ 2 } } =\sqrt { 3 }$

$\Rightarrow { \alpha }^{ 2 }=1\Rightarrow \alpha =\pm 1$

$\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k} , |\vec{b}| = 5$ and the angle between

$\vec{a}$ and

$\vec{b}$ is

$\dfrac{\pi}{6}$ , then the area of the triangle formed by these two vectors as two sides is

0%
$\dfrac{15}{4}$
0%
$\dfrac{15}{2}$
0%
$15$
0%
$\dfrac{15\sqrt 3}{2}$
0%
$15 \sqrt 3$

Explanation

Area of the triangle

$=\dfrac{1}{2} |a \times b|$

$= \dfrac{1}{2} ||a| |b| \sin \theta \widehat n|$

$= \dfrac{1}{2} \left [ 3 \times 5 \times \sin \dfrac{\pi}{6} \right ]$

$\left( \therefore |a| = \sqrt{1 + 2^2 + 2^2 } = 3 \right )$

$= \dfrac{1}{2} \left [ 15 \times \dfrac{1}{2} \right ] = \dfrac{15}{4}$

$\lambda (3 \widehat i + 2 \widehat j - 6 \widehat k)$ is a unit vector, then the values of

$\lambda$ are

0%
$\pm \dfrac{1}{7}$
0%
$\pm 7$
0%
$\pm \sqrt{43}$
0%
$\pm \dfrac{1}{\sqrt{43}}$
0%
$\pm \dfrac{1}{\sqrt{}7}$

Explanation

Now,

$|(3 \widehat i + 2\widehat j - 6\widehat k)| = \sqrt{3^2 + 2^2 + (-6)^2}$

$= \sqrt{9 + 4 + 36}$

$= \sqrt{49} = 7$
Since,

$\lambda (3 \widehat i + 2\widehat j - 6\widehat k)$ is a unit vector.

$\therefore \lambda \pm \dfrac{1}{|3 \widehat i + 2\widehat j - 6\widehat k|}$

$= \pm \dfrac{1}{7}$

Let

$a,b,c$ be three non-zero vectors such that no two of these are collinear. If the vectors

$a+2b$ is collinear with

$c$ and

$b+3c$ is collinear with

$a$ (

$\lambda$ being some non-zero scalar), then

$a+2b+6c$ equals to

0%
$\lambda a$
0%
$\lambda b$
0%
$\lambda c$
0%
$0$

Explanation

Since,

$a+2b$ is collinear with

$c$

$a+2b=mc...(i)$
Since

$b+3c$ is collinear with

$a$

$b+3c=na.....(ii)\quad$
Myltiplying Eq(ii) by

$2$ and subtracting Eq.(ii) from Eq.(i), we get

$a-bc=mc-2na$
On comparing, we get

$m=-6;n=-\cfrac { 1 }{ 2 }$
From Eq.(i), we have

$a+2b=-6c\quad$

$\Rightarrow a+b+6c=0\quad$

Let

$a,b,c$ be three non-zero vectors such that

$a+b+c=0$ , then

$\lambda b\times a+b\times c+c\times a=0$ , where

$\lambda$ is

0%
$1$
0%
$2$
0%
$-1$
0%
$-2$

Explanation

Clearly

$a,b,c$ represents the sides of a triangle.
Therefore,

$a\times b=b\times c=c\times a\quad$

$\Rightarrow a\times b+b\times c+c\times a=3a\times b$

$\Rightarrow 2b\times a+b\times c+c\times a=0$

$\Rightarrow \lambda =2$

Let

$\vec { a } =\hat { i } +\hat { j } -\hat { k } ,\vec { b } =\hat { i } -\hat { j } +\hat { k }$ and

$\vec { c }$ be a unit vector perpendicular to

$\vec { a }$ and coplanar with

$\vec { a }$ and

$\vec { b }$ , then

$\vec { c }$ is

0%
$\cfrac { 1 }{ \sqrt { 2 } } \left( \hat { j } +\hat { k } \right)$
0%
$\cfrac { 1 }{ \sqrt { 2 } } \left( \hat { j } -\hat { k } \right)$
0%
$\cfrac { 1 }{ \sqrt { 6 } } \left( \hat { i } -2\hat { j } +\hat { k } \right)$
0%
$\cfrac { 1 }{ \sqrt { 6 } } \left( 2\hat { i } -\hat { j } +\hat { k } \right)$

Explanation

Required unit vector

$=\pm \cfrac { \vec { a } \times \left( \vec { a } \times \vec { b } \right) }{ \left| \vec { a } \times \left( \vec { a } \times \vec { b } \right) \right| }$
Now

$\vec { a } \times \left( \vec { a } \times \vec { b } \right) =\left( \vec { a } .\vec { b } \right)\vec{a} -\left( \vec { a } .\vec { a } \right) \vec { b }$

$\vec { a } \times \vec { b } =\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix}=-2\hat { j } -2\hat { k }$

$\therefore \vec { a } \times \left( \vec { a } \times \vec { b } \right) =\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & -1 \\ 0 & -2 & -2 \end{vmatrix}=-4\hat { i } +2\hat { j } -2\hat { k }$

$\left| \vec { a } \times \left( \vec { a } \times \vec { b } \right) \right| =\sqrt { 24 } =2\sqrt { 6 }$

$\therefore$ Unit vector

$=\pm \cfrac { \vec { a } \times \left( \vec { a } \times \vec { b } \right) }{ \left| \vec { a } \times \left( \vec { a } \times \vec { b } \right) \right| } =\pm \cfrac { \left( 2\hat { i } -\hat { j } +\hat { k } \right) }{ \sqrt { 6 } }$
(taking positive for option)

$a=2i+2j+3k, b=-1+2j+k$ and

$c = 3i + j$ , then

$a+tb$ this perpendicular to

$c$ ; if

$t$ is equal to

0%
2
0%
4
0%
6
0%
8

Explanation

Given,

$a=2i+2j+3k,b=-1+2j+k$ and

$c=3i+j$

We need to find value of

$a+ tb\perp c$

$\Rightarrow (a+tb)\cdot c=0$

$\Rightarrow a\cdot c+ tb \cdot c=0$

$\Rightarrow t=-\dfrac {a\cdot c}{b\cdot c}$

$\Rightarrow t=-\dfrac {(2i+2j+3k)\cdot (3i+j)}{(-i+2j+k)\cdot (3i+j)}$

$\Rightarrow t=-\dfrac {6+2+0}{-3+2+0}=8$

$3p + 2q =i + j + k$ and

$3p - 2q = i - j - k$ , then the angle between

$p$ and

$q$ is

0%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{2}$
0%
$\pi$

Explanation

Given,

$3p + 2q = i + j + k$ ..... (i)
and

$3p - 2q = i - j - k$ .... (ii)
On adding Eqs. (i) and (ii), we get

$6p = 2i$
Therefore,

$p = \dfrac {i}{3}$
On subtracting Eq. (ii) from Eq. (i), we get

$4q - 2(j + k) \Rightarrow q = \dfrac {1}{2} (j + k)$
Let

$\theta$ be the angle between

$p$ and

$q$ , then

$p\cdot q = |p||q|\cos \theta$

$\Rightarrow \cos \theta = \dfrac {p\cdot q}{|p||q|}$

$\Rightarrow \cos \theta = \dfrac {1}{6} \dfrac {i\cdot (j + k)}{|p||q|} = 0 = \cos 90^{\circ}$
Thus

$\theta = 90^{\circ} = \dfrac {\pi}{2}$

$\overline{a},\overline{b},\overline{c},$ are unit vectors such that

$\overline{a}+\overline{b}+\overline{c}+\overline{c.a}=$

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

Explanation

$-\dfrac{3}{2}$

If the scalar projection of the vectors

$xi-j+k$ on the vector

$2i-j+5k$ is

$\cfrac { 1 }{ \sqrt { 30 } }$ , then the value of

$x$ is equal to

0%
$-\cfrac { 5 }{ 2 }$
0%
$6$
0%
$-6$
0%
$3$

Explanation

Projection of

$xi-j+k$ on

$2i-j+5k$

$=\cfrac { \left( xi-j+k \right) \left( 2i-j+5k \right) }{ \sqrt { 4+1+25 } } =\cfrac { 2x+1+5 }{ \sqrt { 30 } }$
But

$\cfrac { 2x+6 }{ \sqrt { 30 } } =\cfrac { 1 }{ \sqrt { 30 } }$

$\Rightarrow x=-\cfrac { 5 }{ 2 }$

$a$ and

$c$ are unit vectors and

$|b| = 4$ . If angle between

$b$ and

$c$ is

$\cos^{-1}\left (\dfrac {1}{4}\right )$ and

$a\times b = 2a\times c$ , then

$b= \lambda a + 2c$ , where

$\lambda$ is equal to

0%
$\pm \dfrac {1}{4}$
0%
$\pm \dfrac {1}{2}$
0%
$\pm 4$
0%
None of the above

Explanation

Given,

$a\times b = 2a\times c$

$\Rightarrow a\times b - 2a\times c = 0$

$\Rightarrow a\times (b - 2c) = 0$

$\Rightarrow a$ and

$(b - c)$ are collinear.

$\Rightarrow b - 2c = \lambda a$

$\Rightarrow |b - 2c|^{2} = \lambda^{2}|a|^{2}$

$\Rightarrow |b|^{2} + 4|c|^{2} - 4b\cdot c = \lambda^{2}|a|^{2}$

$\Rightarrow 16 + 4 - 4 (4) (1)\dfrac {1}{4} = \lambda^{2}(1)$
Therefore,

$\lambda = \pm 4$ .

Let

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

$\vec{OA} =4\hat { i } +2\hat { j } +2\hat { k }$ . The distance of the point

$B$ from the straight line passing through

$A$ and parallel to the vector

$2\hat { i } +3\hat { j } +6\hat { k }$ is

0%
$\dfrac { 7\sqrt { 5 } }{ 9 }$
0%
$\dfrac { 5\sqrt { 7 } }{ 9 }$
0%
$\dfrac { 3\sqrt { 5 } }{ 7 }$
0%
$\dfrac { 9\sqrt { 5 } }{ 7 }$
0%
$\dfrac { 9\sqrt { 7 } }{ 5 }$

Explanation

Let the equation of line be

$r=a+\lambda b$ .

Since, line passes through

$A$ , so

$a=4\hat { i } +2\hat { j } +2\hat { k }$ and line is parallel to the vector

$2\hat { i } +3\hat { j } +6\hat { k }$ , so

$b=2\hat { i } +3\hat { j } +6\hat { k }$
Hence, equation of the line is

$r=4\hat { i } +2\hat { j } +2\hat { k } +\lambda \left( 2\hat { i } +3\hat { j } +6\hat { k } \right)$
Distance of point

$B$ from the line

$=\left| \dfrac { \left| \left( { a }_{ 2 }-{ a }_{ 1 } \right) \right| \times b }{ \left| b \right| } \right|$

${ a }_{ 2 }-{ a }_{ 1 }=3\hat { i } +0\hat { j } +0\hat { k }$

$\left( { a }_{ 2 }-{ a }_{ 1 } \right) \times b=\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 0 & 0 \\ 2 & 3 & 6 \end{vmatrix}=-18\hat { j } +9\hat { k }$

$\left| \left( { a }_{ 2 }-{ a }_{ 1 } \right) \times b \right| =\sqrt { { \left( -18 \right) }^{ 2 }+{ \left( 9 \right) }^{ 2 } }$

$=\sqrt { 405 } =9\sqrt { 5 }$

$\left| b \right| =\sqrt { 4+9+36 } =7$
Therefore, required distance is

$\dfrac { 9\sqrt { 5 } }{ 7 }$ .

Given that A

$+$ B

$+$ C

$=0$ . Out of three vectors, two are equal in magnitude and the magnitude of third vector is

$\sqrt{2}$ times that of either of the two having equal magnitude. Then, the angles between the vectors are given by.

0%
$30^o, 60^o, 90^o$
0%
$45^o, 45^o, 90^o$
0%
$45^o, 60^o, 90^o$
0%
$90^o, 135^o, 135^o$

Explanation

From polygon law, three vectors having summation zero should form, a closed polygon(triangle). Since, the two vectors are having same magnitude and the third vector is

$\sqrt{2}$ times that of either of two having equal magnitude, i.e., the triangle should be right angled triangle.
Angle between A and B,

$\alpha =90^o$
Angle between B and C,

$\beta =135^o$
Angle between A and C,

$\gamma =135^o$ .

The vectors

$2\hat{i} + 3\hat{j}$ ,

$5\hat{i}+ 6\hat{j }$ and

$8\hat{i }+ \lambda\hat{j }$ have their initial points at ( 1, 1 ) . Find the value of

$\lambda$ so that the vectors terminate on one straight line.

0%
9
0%
8
0%
7
0%
6

$\overline {e} = l\overline {i} + m\overline {j} + n\overline {k}$ is a unit vector, the maximum value of

$lm + mn + nl$ is

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

Explanation

Given

$\vec{e}$ is unit vector,

$\therefore l^2+m^2+n^2=1\dots \dots(1)$

We know that the square of any real number is always greater than zero.

$\Rightarrow (l+m+n)^2\ge0$

$\Rightarrow l^2+m^2+n^2+2(lm+mn+nl)\ge 0$

From

$(1)$ , we get

$lm+nm+nl\ge-\dfrac{1}{2}$

$|\vec {a}| = 3, |\vec {b}| = 4$ and

$|\vec {a} - \vec {b}| = 7$ then

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

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

We know that

$\Rightarrow| \vec{a}-\vec{b} |^2=| \vec{a} |^2 +| \vec{b} |^2-2| \vec{a} || \vec{b} |\cos{\theta}$

$\Rightarrow 7^2=3^2+4^2-2(3)(4)\cos{\theta}$

$\Rightarrow 49-9-16=-24\cos{\theta}$

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

$\Rightarrow| \vec{a}+\vec{b} |^2=| \vec{a} |^2 +| \vec{b} |^2+2| \vec{a} || \vec{b} |\cos{\theta}$

$\Rightarrow| \vec{a}+\vec{b} |^2=3^2 +4^2+2(3)(4)(-1)$

$\Rightarrow| \vec{a}+\vec{b} |^2=9+16-24$

$\Rightarrow| \vec{a}+\vec{b} |^2=1$

$\Rightarrow| \vec{a}+\vec{b} |=1$

Hence, answer is option

$A$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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