If $$\vec{a}.\vec{b}=0$$ and $$\vec{a}\times \vec{b}=0$$ then

$$\vec{a}=\vec{0}$$ or $$\vec{b}=\vec{0}$$

MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 5

A man starts from

$O$ moves

$500m$ turns by

$60 ^\circ$ and moves

$500m$ again turns by

$60 ^\circ$ and moves

$500m$ and so on. Find the displacement after

$(i)$ 5th turn ,

$(ii)$ 3rd turn.

0%
- $500m, 1000m$
0%
$500m, 500 \sqrt {3} m$
0%
$1000m, 500 \sqrt {3} m$
0%
none of these

Explanation

(i) Fifth turn is at E and this displacement will be given by vector

$\vec{OE}$ which is clear from the given fig that it is-

$500m$ .

(ii) Third turn is at C and this will be given by vector

$\vec{OC}$ which is equal to

$\sqrt{AC^2+AB^2}=\sqrt{(500\sqrt{3})^2+500^2}=1000m$ where

$AC=2AB \ sin60^o$

The projection of the vector

$\hat i - 2\hat j + \hat k$ on the vector

$4\hat i - 4\hat j + 7\hat k$ is

0%
$\displaystyle \dfrac{5\sqrt{6}}{10}$
0%
$\displaystyle \dfrac{19}{9}$
0%
$\displaystyle \dfrac{9}{19}$
0%
$\displaystyle \dfrac{\sqrt{6}}{19}$

Explanation

Projection of

$\vec a$ on

$\vec b$

$= \displaystyle |a| \cos \theta = |a| \dfrac{a \cdot b}{|a||a|} = \dfrac{a \cdot b}{|b|}$

$= \displaystyle \dfrac{4 + 8 + 7}{\sqrt{16 + 16 + 49}}$

$= \dfrac{19}{\sqrt{81}}$

$= \dfrac{19}{9}$

If the vector product of a constant vector

$\displaystyle \vec{OA}$ with a variable vector

$\displaystyle \vec{OB}$ in a fixed plane

$OAB$ be a constant vector, then locus of

$B$ is :

0%
A straight line perpendicular to $\displaystyle \vec{OA}$
0%
A circle with centre O radius equal to $\displaystyle \left | \vec{OA} \right |$
0%
A straight line parallel to $\displaystyle \vec{OA}$
0%
None of these

Explanation

Vector product of a vector with some vector in the plane, leads a constant vector; implies that the answer can only be

$0$ .
Thus, we will have a vector parallel to the original vector so that the angle between the two becomes

$0$ and the answer comes out to be

$0$ .

There are

$N$ co-planar vectors each of magnitude

$V$ . Each vector is inclined to the preceding vector at angle

$2 \pi/N$ . What is the magnitude of their resultant?

0%
zero
0%
$V/N$
0%
$V$
0%
$NV$

Explanation

When vectors are arranges at an angle

$\dfrac { 2\pi }{ N }$ to the preceding vectors, they form an N-sided closed polygon. Adding all the sides of a polygon we get the resultant to be 0.

$P, Q$ have position vectors

$\vec {a}$ &

$\vec {b}$ relative to the origin '

$O$ ' and

$X, Y$ divide

$\vec{PQ}$ intermally and extemally respectively in the ratio

$2:1$ , then vector

$\vec{XY}=$

0%
$\displaystyle \:\dfrac{3}{2}(\vec b - \vec a)$
0%
$\displaystyle \:\dfrac{4}{3}(\vec c - \vec a)$
0%
$\displaystyle \:\dfrac{5}{6}\left ( \vec{b}- \vec{a}\right )$
0%
$\displaystyle \:\dfrac{4}{3}(\vec b - \vec a)$

Explanation

$X$ divides

$\vec{PQ}$ internally in

$2:1$ . Thus,

$\vec X = \dfrac{\vec a + 2\vec b}{2 + 1}$

$Y$ divides

$\vec{PQ}$ externally in

$2:1$ . Thus,

$\vec Y = \dfrac{2\vec b - \vec a}{2 - 1}$

$\vec X = \dfrac{\vec a + 2\vec b}{3}$ and

$\vec Y = 2\vec b - \vec a$

$\therefore \vec{XY} = \vec Y - \vec X = \dfrac{6\vec b - 3\vec a -\vec a -2\vec b}{3}$

$\vec{XY}= \dfrac{4\vec b - 4\vec a}{3}$

If . and

$\times$ represent dot product and cross product respectively then which of the following is meaningless?

0%
$\left ( \vec{a}\times \vec{b} \right ).\left ( \vec{c}\times \vec{d} \right )$
0%
$\left ( \vec{a}\times \vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$
0%
$\left ( \vec{a}.\vec{b} \right ) .\left ( \vec{c}\times \vec{d} \right )$
0%
$\left ( \vec{a}.\vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$

Explanation

Cross product can only occur for two vectors.

$a.b$ is just a scalar quantity.
Hence the option

$D$ is meaningless.

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

$\vec{a}\times \vec{b}=0$ then

0%
$\vec{a}\parallel \vec{b}$
0%
$\vec{a}\perp \vec{b}$
0%
$\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$
0%
None of these

Explanation

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

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

$\vec{a}.\vec{b} = ab \cos{\theta}$ and

$|\vec{a} \times \vec{b}| = ab \sin{\theta}$

Since both of them are simultaneously

$0$ and

$\cos{\theta}, \sin{\theta}$ cannot be simultaneously

$0$ , we get either

$a$ or

$b$ has to be

$0$ .

$G$ and

$G'$ be the centroids of the triangles

$ABC$ and

$A'B'C'$ respectively, then

$\overrightarrow { AA' } +\overrightarrow { BB' } +\overrightarrow { CC' } =$

0%
$\displaystyle \dfrac { 2 }{ 3 } \overrightarrow { GG' }$
0%
$\overrightarrow { GG' }$
0%
$2\overrightarrow { GG' }$
0%
$3\overrightarrow { GG' }$

Explanation

Let P.V. of

$A\left( \overrightarrow { a } \right) ,B\left( \overrightarrow { b } \right) ,C\left( \overrightarrow { c } \right) ,A'\left( \overrightarrow { { a }_{ 1 } } \right) ,B'\left( \overrightarrow { { b }_{ 1 } } \right) ,C'\left( \overrightarrow { { c }_{ 1 } } \right)$

Now P.V. of

$\displaystyle G=\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 }$ ;

Similarly

$\displaystyle G'=\dfrac { \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } }{ 3 }$

Now

$\overrightarrow { AA' } =\overrightarrow { { a }_{ 1 } } -\overrightarrow { a } ,\overrightarrow { BB' } =\overrightarrow { { b }_{ 1 } } -\overrightarrow { b } ,\overrightarrow { CC' } =\overrightarrow { { c }_{ 1 } } -\overrightarrow { c }$

$\overrightarrow { AA' } +\overrightarrow { BB' } +\overrightarrow { CC' } =\left( \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } \right) -\left( \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } \right)$

$\displaystyle =3\left( \dfrac { \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } }{ 3 } -\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } \right) =3\overrightarrow { GG' }$

Five coplanar forces of equal magnitudes

$10 N$ each, act at a point such that the angle between any two consecutive forces is same. The magnitude of their resultant is :

0%
$0$
0%
$10 N$
0%
$20 N$
0%
$10 \sqrt {2} N$

Explanation

As the angle between any two consecutive forces have the same value

${ =72}^{\circ}$ and all the forces are of equal magnitude. Hence their resultant is zero.

$P$ is a point on the line through the point

$A$ whose position vector is

$\overrightarrow{a}$ and the line is parallel to the vector

$\overrightarrow{b}$ . If

$PA=6$ , the position vector of

$P$ is

0%
$\overrightarrow{a}+6\overrightarrow{b}$
0%
$\displaystyle \overrightarrow{a}+\dfrac{6}{\left |\overrightarrow{b} \right |}\overrightarrow{b}$
0%
$\overrightarrow{a}-6\overrightarrow{b}$
0%
$\displaystyle \overrightarrow{b}+\dfrac{6}{\left |\overrightarrow{a} \right |}\overrightarrow{a}$

Explanation

Line is parallel to vector

$\overrightarrow{b}$
The required vector is at a distance of 6 units from

$\overrightarrow{a}$
So, the required position vector will be

$\overrightarrow{a} + \dfrac{6}{|\overrightarrow{b}|} \overrightarrow{b}$ since we need a unit vector in the direction of

$\overrightarrow{b}$

A straight line

$r=a+\lambda b$ meets the plane

$r.n=0$ in

$P$ . The position vector of

$P$ is

0%
$\displaystyle a+\frac { a.n }{ b.n } b$
0%
$\displaystyle a+\frac { b.n }{ a.n } b$
0%
$\displaystyle a-\frac { a.n }{ b.n } b$
0%
None of these

Explanation

A straight line

$r=a+\lambda b$ meets the plane

$r.n=0$ in

$P$ for which

$\lambda$ is given by

$\displaystyle \left( a+\lambda b \right) .n=0\Rightarrow \lambda =-\frac { a.n }{ b.n }$

Thus, the position vector of

$P$ is

$\displaystyle r=a-\frac { a.n }{ b.n } b$

$[$ putting the value of

$\lambda$ in

$r=a+\lambda b]$

Let

$\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$ and

$\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$ . If the point

$P$ on the line segment

$BC$ is equidistant from

$AB$ and

$AC$ then

$\overrightarrow{AP}$ is

0%
$2\hat{i}-\hat{k}$
0%
$\hat{i}-2\hat{k}$
0%
$2\hat{i}+\hat{k}$
0%
None of these

Explanation

Since

$P$ is equidistant from

$AC\ \&\ BC,$ it lies on the angle bisector of

$\angle{BAC}$
Let us also assume that

$A$ is the origin.

So,

$\overrightarrow{P} = x\dfrac{3\hat{i} + \hat{j} - \hat{k} + \hat{i} - \hat{j} + 3\hat{k}}{\sqrt{11}}$

$\Rightarrow \overrightarrow{P} = y{4\hat{i} + 2\hat{k}}$

Hence we can see the options to verify.

If the vectors

$\vec{a}$ ,

$\vec{b}$ ,

$\vec{c}$ and

$\vec{d}$ are coplanar, then

$\left ( \vec{a}\times \vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$ is equal to

0%
$\vec{a}+\vec{b}+\vec{c}+\vec{d}$
0%
$\vec{0}$
0%
$\vec{a}+\vec{b}=\vec{c}+\vec{d}$
0%
None of these

Explanation

Since

$\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar, the cross products

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

$\vec{c} \times \vec{d}$ is a vector same for both.

$\Rightarrow$ their cross product will become 0 since the sine of angle between becomes 0.

$\vec{a}$ ,

$\vec{b}$ ,

$\vec{c}$ ,

$\vec{d}$ are any four vectors then

$\left ( \vec{a}\times \vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$ is a vector

0%
Perpendicular to $\vec{a}$ , $\vec{b}$ , $\vec{c}$ , $\vec{d}$
0%
Along the line of intersection of two planes, one containing $\vec{a}$ , $\vec{b}$ and the other containing $\vec{c}$ , $\vec{d}$
0%
Equally inclined to both $\vec{a}\times \vec{b}$ and $\vec{c}\times \vec{d}$
0%
None of these

Explanation

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

checking for option (A)

if dot product with

$\vec{a},\vec{b},\vec{c}\ and\ \vec{d}$ gives 0 then correct or perpendicular

so taking dot product with

$\vec{a}$

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

$[\vec{a}\vec{b}\vec{c}]\times(\vec{c}\times\vec{d})\neq0$

similarly invalid for vectors b,c,d

Cheking option (b)

eq of line of intersection of two planes

$\vec{r}=(x\hat{i}+y\hat{j}+z\hat{k})+\lambda(\vec{n_{1}}\times\vec{n_{2}})$

where

$\vec{n_{1}},\vec{n_{2}}$ are normal vector of planes

SO if planes

$P_{1}\ and\ P_{2}$ contains

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

then

$\vec{n_{1}}\times\vec{n_{2}}=(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})$

IT is valid

SO third option also valid above solution

The position vectors of two vertices and the centroid of a triangle are

$\vec{i}+\vec{j}$ ,

$2\vec{i}-\vec{j}+\vec{k}$ and

$\vec{k}$ respectively, then the position vector of the third vertex of the triangle is

0%
$-3\vec{i}+2\vec{k}$
0%
$3\vec{i}-2\vec{k}$
0%
$\vec{i}+\dfrac{2}{3}\vec{k}$
0%
None of these

Explanation

Let the third vertex be

$C=x\vec{i}+y\vec{j}+z\vec{k}$
and

$A=(1,1,0)$ ,

$B=(2,-1,1)$ ,

$G=(0,0,1)$
Hence

$G=\left (\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3},\dfrac{z_{1}+z_{2}+z_{3}}{3}\right)$

$(0,0,1)=\left (\dfrac{1+2+x}{3},\dfrac{1-1+y}{3},\dfrac{0+1+z}{3}\right)$

$(0,0,1)=\left (\dfrac{x+3}{3},\dfrac{y}{3},\dfrac{z+1}{3}\right)$
Hence

$x+3=0$ gives

$x=-3$

$\dfrac{y}{3}=0$ gives

$y=0$
And

$\dfrac{z+1}{3}=1$ gives

$z=2$
Hence

$C=-3\vec{i}+2\vec{k}$ .

$\displaystyle R(\bar{r})$ is any point on a semi-circle,

$\displaystyle P(\bar{p})$ and

$\displaystyle Q(\bar{q})$ are the position vector of the end point of the diameter of that semi-circle, then

$\displaystyle \overline{PR}\cdot \overline{QR}$ is equal to

0%
$1$
0%
$0$
0%
$3$
0%
Not defined

Explanation

If any point on the semi-circle lies and we join that point to the end points of diameter then the angle formed between them is always normal or 90

So the angle between

$\vec{PR}\ and\ \vec{QR}$ be 90

$\vec{PR}\cdot\vec{QR}=\left |\vec{PR}\right | \left |\vec{QR}\right |\cos90$

$\vec{PR}\cdot\vec{QR}=\left |\vec{PR}\right | \left |\vec{QR}\right |\times0$

$\vec{PR}\cdot\vec{QR}=0$

In a triangle OAB, E is the mid-point of OB and D is a point on AB such that AD: DB = 2 :If OD and AE intersect at P, determine the ratio OP : PD using vector methods.

0%
$OP:PD=2:3$
0%
$OP:PD=3:2$
0%
$OP:PD=1:3$
0%
$OP:PD=3:1$

Explanation

With

$O$ as origin let

$a$ and

$b$ be the position vectors of

$A$ and

$B$ respectively.
Then the position vector of

$E$ , the mid-point of

$OB$ , is

$\displaystyle\frac{b}{2}$
Again, since

$AD: DB=2:1$ , the position vector of

$D$ is

$\displaystyle \frac{1\cdot a+2b}{1+2}=\frac{a+2b}{3}$
Equation of

$OD$ and

$AE$ are

$\displaystyle r=t\frac{a+2b}{3}$ ...(1)

and

$\displaystyle r=a+s\left ( \frac{b}{2}-a \right )$ or

$\displaystyle r=\left ( 1-s \right )a+s\frac{b}{2}$ ...(2)
If they intersect at

$p$ , then we will have identical values of

$r$ .

Hence comparing the coefficients of

$a$ and

$b$ , we get

$\displaystyle \frac{t}{3}=1-s,\frac{2t}{3}=\frac{s}{2}\therefore t=\frac{3}{5}$ or

$\displaystyle s=\frac{4}{5}.$
Putting for t in (1) or for s in (2), we get the position vector of point of intersection

$P$ as

$\displaystyle \frac{a+2b}{5}$ ...(3)
Now let

$P$ divide

$OD$ in the ratio

$\displaystyle\lambda :1.$

Hence by ratio formula the

$P.V.$ of

$P$ is

$\displaystyle \dfrac{\dfrac{\lambda \left ( a+2b \right )}{3}+1.0}{\lambda +1}=\dfrac{\lambda }{3\left(\lambda +1 \right )}\left ( a+2b \right )$ ....(4)
Comparing (3) and (4), we get

$\displaystyle \frac{\lambda }{3\left ( \lambda +1 \right )}=\frac{1}{5}$

$\Rightarrow\displaystyle 5\lambda =3\lambda +3\Rightarrow 2\lambda =3$

$\displaystyle\Rightarrow \lambda =\frac{3}{2}$

$\therefore OP:PD=3:2$

In a triangle ABC, D divides BC in the ratio 3 : 2 and E divides CA in the ratio 1 :The lines AD and BE meet at H and CH meets AB in F. Find the ratio in which F divides AB.

0%
$\displaystyle AF:FB=2:1$
0%
$\displaystyle AF:FB=1:2$
0%
$\displaystyle AF:FB=2:3$
0%
$\displaystyle AF:FB=3:2$

Explanation

Take

$A$ as origin and the position vectors of

$B$ and

$C$ be

$b$ and

$c$ .

Hence the position vectors of other points under given conditions are

$\displaystyle D=\dfrac{3c+2b}{5}, E=\dfrac{1.0+3c}{4}= \dfrac{3}{4}c.$

Equations of

$AD$ and

$BE$ are

$AD$ is

$\displaystyle r= t\dfrac{3c+2b}{5}$

$BE$ is

$\displaystyle r= (1- s )b+s\cdot\dfrac{3}{4}c.$

They intersect at

$H$ .

Comparing coefficients of

$b$ and

$c$ , we get

$\displaystyle \dfrac{2}{5}t= 1-s, \dfrac{3}{5}t= \dfrac{3}{4}s.\\$

$\displaystyle \therefore s= \dfrac{4}{5}t.\\$

$\displaystyle\therefore \dfrac{2}{5}t+\dfrac{4}{5}t= 1\\$

$\displaystyle \therefore t= \dfrac{5}{6}, s=\dfrac{4}{6}\\$

Point

$H$ is

$\displaystyle \dfrac{3c+2b}{6}.$

Now

$F$ is point of inersection of

$AB$ and

$CH$ whose equations are

$r = t b$

and

$\displaystyle r=\left ( 1-s \right )c+s\dfrac{3c+2b}{6}$

Comparing the coefficients,

$\displaystyle t=\dfrac{2s}{6}=\dfrac{s}{3}$ and

$\displaystyle 1-s+\dfrac{3}{6}s=0\Rightarrow s=2\Rightarrow t=\dfrac{1}{3}s=\dfrac{2}{3}$

$\displaystyle \therefore$ P.V. of

$\displaystyle F=\dfrac{2}{3}b$ or

$\displaystyle \vec{AF}=\dfrac{2}{3}b,\vec{FB}=b-\dfrac{2}{3}b=\dfrac{1}{3}b$

$\therefore AF:FB=2:1$

$\displaystyle \bar{a}= x\hat{i}+y\hat{j}+z\hat{k},\bar{b}= \hat{j}$ then the vector

$\displaystyle \bar{c}$ for which

$\displaystyle \bar{a},\bar{b},\bar{c}$ form a right hand triad

0%
$\displaystyle x(\hat{i}-\hat{k})$
0%
$\displaystyle \bar{0}$
0%
$\displaystyle -z\hat{i}+x\hat{k}$
0%
$\displaystyle y\hat{j}$

Explanation

Three vectors a, b and c are said to form a right handed system if a right threaded screw when rotated from a to b the screw moves in direction of c.

Even you can say that axb=c

$\vec{c}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k} \\x&y&z\\0&1&0\end{vmatrix}$

$\vec{c}=\hat{i}(-z)-\hat{j}(0)+\hat{k}(x)$

$\vec{c}=-z\hat{i}+x\hat{k}$

Let

$\displaystyle \bar{p}$ and

$\displaystyle \bar{q}$ be two distinct points. Let

$R$ and

$S$ be the points dividing

$PQ$ internally and externally in the ratio

$2:3$ . If

$\displaystyle \overline{OR} \perp \overline{OS},$ then

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

Explanation

Position vector of

$\displaystyle R=\dfrac{2\bar{q+3\bar{p}}}{5}$
position vector of

$\displaystyle S=\dfrac{2\bar{q-3\bar{p}}}{\left ( -1 \right )}$

$\displaystyle =3\vec{p}-2\bar{q}$
Since

$\displaystyle \overline{OR}\perp \overline{OS}$ ,

$\therefore \overline{OR}\cdot \overline{OS}=0$

$\displaystyle \therefore 9p^{2}=4q^{2}$

The difference of the squares on the diagonals is four times the rectangle contained by either of these sides and the projection of the other upon it.

0%
$\displaystyle = 4$ ( rectangle contained by AB and projection of AC on AB).
0%
$\displaystyle = 1$ ( rectangle contained by AB and projection of AC on AB).
0%
$\displaystyle = 8$ ( rectangle contained by AB and projection of AC on AB).
0%
$\displaystyle = 2$ ( rectangle contained by AB and projection of AC on AB).

Explanation

Again

$\displaystyle \vec{AD}^{2}-\vec{BC}^{2}= \left ( b+c \right )^{2}-\left ( c-b \right )^{2}= 4b.c$

$\displaystyle = 4\vec{AB}.\vec{AC}= 4AB$

$($ Projection of

$AC$ on

$AB)$

$\displaystyle = 4$

$($ rectangle contained by

$AB$ and projection of

$AC$ on

$AB).$

State true or false:

The four diagonals of a parallelopiped and the joins of the mid-points of opposite edges are concurrent at a common point of bisection.

0%
True
0%
False

$ABC$ is a

$\displaystyle \Delta$ and

$G$ is its centroid. If

$\displaystyle \overline{AB}= \bar{b}$ and

$\displaystyle \overline{AC}= \bar{c}$ , then

$\displaystyle \overline{AG}$ is equal to

0%
$\displaystyle \dfrac{2}{3}\left ( \bar{b}+\bar{c} \right )$
0%
$\displaystyle \bar{b}+\bar{c}$
0%
$\displaystyle \dfrac{1}{3}\left ( \bar{b}+\bar{c} \right )$
0%
$\displaystyle \dfrac{3}{2}\left ( \bar{b}+\bar{c} \right )$

Explanation

Taking

$A$ as origin.
So, position vector of

$\displaystyle B=\overline{AB}=\bar{b}$
and position vector of

$\displaystyle C=\overline{AC}=\bar{c}$
Hence by centroid formula,

$\displaystyle \overline{AG}$ = position vector of

$G$

$\displaystyle =\dfrac{\bar{a} + \bar{b}+\bar{c}}{3}$

$= \dfrac{\bar{b}+\bar{c}}{3}$

In a parallelogram

$ABCD,\left| AB \right| =a,\left| AD \right| =b$ and

$\left| AC \right| =c.$ Then,

$DB.AB$ has the value

0%
$\displaystyle\frac { 3{ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$
0%
$\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$
0%
$\displaystyle\frac { { a }^{ 2 }-{ b }^{ 2 }+3{ c }^{ 2 } }{ 2 }$
0%
$\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }+{ c }^{ 2 } }{ 2 }$

Explanation

$\because DB=DA+AB$ or,

$DA=DB-AB$

$\displaystyle \therefore { \left( DA \right) }^{ 2 }={ \left( DB \right) }^{ 2 }+{ \left( AB \right) }^{ 2 }-2DB.AB$

In parallelogram

$\displaystyle 2\left( { a }^{ 2 }+{ b }^{ 2 } \right) ={ c }^{ 2 }+{ DB }^{ 2 }$

$\displaystyle \therefore { \left( DB \right) }^{ 2 }=2{ a }^{ 2 }+2{ b }^{ 2 }-{ c }^{ 2 }$

$\therefore$ From (1),

$\displaystyle { b }^{ 2 }=2{ a }^{ 2 }+{ 2b }^{ 2 }-{ c }^{ 2 }+{ a }^{ 2 }-2AB.DB$

$\displaystyle \therefore AB.DB=\dfrac { 3{ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$ .

Let

$\bar{a}$ ,

$\bar{b}$ and

$\bar{c}$ be vectors with magnitudes

$3, 4$ and

$5$ respectively and

$\bar{a}+\bar{b}+\bar{c}=0$ , then the value of

$\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}$ is

0%
$48$
0%
$-26$
0%
$25$
0%
$-25$

Explanation

We have

$|a|= 3$

$|b|=4$

$|c|=5$
Clearly (as

$c^{2}=a^{2}+b^{2}$ )

$\bar{a}$ ,

$\bar{b}$ and

$\bar{c}$ forms the right angle

$\triangle$

$\therefore$

$\bar{a}.\bar{b}=0$

$\therefore$

$\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}=\bar{c}.\left ( \bar{a}+\bar{b} \right )$

$=\bar{c}.\left ( -\bar{c} \right )$

$=-\left | \bar{c} \right |^{2}=-25$

The sides of a

$\triangle$ are in A.P, then the line joining the centroid to the incenter is parallel to

0%
The largest side
0%
The middle side
0%
The smallest side
0%
None of these

Explanation

$G=(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3})$

$=(\dfrac{bx_{1}+bx_{2}+bx_{3}}{3b},\dfrac{by_{1}+by_{2}+by_{3}}{3b})$

$I=(\dfrac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\dfrac{ay_{1}+by_{2}+cy_{3}}{a+b+c})$

$2b=(a+c)$

Hence

$I=(\dfrac{ax_{1}+bx_{2}+cx_{3}}{3b},\dfrac{ay_{1}+by_{2}+cy_{3}}{3b})$

$\vec{IG}=\dfrac{(b-a)x_{1}+(b-c)x_{2}}{3b}i+\dfrac{(b-a)y_{1}+(b-c)y_{2}}{3b}j$

$A=(x_{1},y_{1})$

$B=(x_{2},y_{2})$

$C=(x_{3},y_{3})$

Then we can see that

$\vec{IG}$ is parallel to the intermediate side.

$\bar{\alpha }$ ,

$\bar{\beta }$ and

$\bar{\gamma }$ be vertices of a

$\triangle$ whose circumcenter is at the origin, then orthocenter is given by

0%
$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{4}$
0%
$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{2}$
0%
$\bar{\alpha }+\bar{\beta }+\bar{\gamma }$
0%
$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{3}$

Explanation

We have

$\bar\alpha$ ,

$\bar\beta$ ,

$\bar\gamma$ vertices of a triangle.
Centroid of the

$\triangle$ is

$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{3}$
Now centroid divides the line joining orthocenter and circumcenter in ratio

$2:1$

$\therefore$ orthocenter of the

$\triangle$ is

$\bar{\alpha }+\bar{\beta }+\bar{\gamma }$

$S$ is the circumcenter,

$O$ is the orthocenter of

$\triangle ABC$ , then

$\displaystyle \vec{SA}+\vec{SB}+\vec{SC}=$

0%
$2\vec{OS}$
0%
$2\vec{SO}$
0%
$\vec{OS}$
0%
$\vec{SO}$

Explanation

A very important property to be remembered of a triangle is that the centroid divides the line joining orthocentre and circumcentre in the ratio

$2 : 1$

$\Rightarrow C = \dfrac{2S + O}{3}$

$\Rightarrow\dfrac{\overline{A} + \overline{B} + \overline{C}}{3} = \dfrac{2\overline{S} + \overline{O}}{3}$

$\Rightarrow \overline{A} + \overline{B} + \overline{C} = 2\overline{S} + \overline{O}$

Thus,

$\overline{SA} + \overline{SB} + \overline{SC} = \overline{A} + \overline{B} + \overline{C} - 3\overline{S} = \overline{SO}$

For non-zero vectors

$\bar{a}$ ,

$\bar{b}$ and

$\bar{c}$ ,

$\left | \left ( \bar{a}\times \bar{b} \right ).\bar{c} \right |=\left | \bar{a} \right |\left | \bar{b} \right |\left | \bar{c} \right |$ iff

0%
$\bar{a}.\bar{c}=0$ , $\bar{a}.\bar{b}=0$
0%
$\bar{a}.\bar{b}=0$ , $\bar{b}.\bar{c}=0$
0%
$\bar{c}.\bar{a}=0$ , $\bar{b}.\bar{c}=0$
0%
$\bar{a}.\bar{b}=\bar{b}.\bar{c}=\bar{c}.\bar{a}=0$

Explanation

$\left | \left ( \overrightarrow{a}\times \overrightarrow{b} \right ).\overrightarrow{c} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\left | \overrightarrow{c} \right |$

$(\theta$ is angle between

$\overrightarrow{c}$ and plane of

$\overrightarrow{a}\times \overrightarrow{b})$

$\Rightarrow \overrightarrow{c}$ is

$\perp$ to

$\overrightarrow{a}$ and

$\overrightarrow{b}$ for

$\cos \theta =1$

and

$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |$ iff

$\overrightarrow{a}\perp \overrightarrow{b}$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=0=\overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{c}\cdot \overrightarrow{a}$

Two planes are perpendicular to each other,one of them contains vector

$\bar{a}$ and

$\bar{b}$ , other contains

$\bar{c}$ and

$\bar{d}$ then

$\left ( \bar{a}\times \bar{b} \right )\cdot \left (\bar{c}\times \bar{d} \right )=$

0%
$1$
0%
$0$
0%
$\left [ \bar{a}\bar{b}\bar{c} \right ]$
0%
$\left [ \bar{b}\bar{c}\bar{d} \right ]$

Explanation

Normal vector to plane

$\bar{a}, \bar{b}$ is given by,

$\bar{a}\times \bar{b}$
And normal vector to plane

$\bar{c}, \bar{d}$ is given by,

$\bar{c}\times \bar{d}$
Now for planes to be pependicular, the angle between their normal should be

$90^{\circ}$

$\Rightarrow (\bar{a}\times \bar{b})\cdot (\bar{c}\times \bar{d}) = |\bar{a}\times \bar{b}||\bar{c}\times \bar{d}|\cos 90^{\circ} = 0$

Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then

$\bar{OA}+\bar{OB}+\bar{OC}+\bar{OD}=$

0%
$2\bar{OP}$
0%
$3\bar{OP}$
0%
$\bar{OP}$
0%
$4\bar{OP}$

Explanation

Let the position vectors of

$A, B, C, D, P$ be

$\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{p}$ respectively.
We need

$\overline{a} + \overline{b} + \overline{c} + \overline{d}$
Also,

$\overline{p} = \dfrac{\overline{a} + \overline{b}}{2} = \dfrac{\overline{c} + \overline{d}}{2}$ .........

$(\because$ diagonals bisect each other

$)$
So, the question needed becomes

$4\overline{p} = 4\overline{OP}$

If a.

$\displaystyle b\neq 0,$ find the vector r which satisfies the equations

$\displaystyle \left ( r-c \right )\times b= 0, r.a= 0$

0%
$\displaystyle r= \frac{\left ( a.b \right )c+\left ( a.c \right )b}{a.b}$
0%
$\displaystyle r= \frac{\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$
0%
$\displaystyle r= \frac{-\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$
0%
$\displaystyle r= \frac{-\left ( a.b \right )c+\left ( a.c \right )b}{a.b}$

Explanation

$\displaystyle \left ( r-c \right )\times b= 0\Rightarrow r-c$ and b are collinear

$\displaystyle \therefore r-c= kb$ or

$\displaystyle r= c+kb$ ...(1)

Again

$\displaystyle r.a=0 \therefore \left ( c+kb \right ).a= 0$ or

$\displaystyle c.a+kb.a= 0 \therefore k= -\dfrac{a.c}{a.b}$

Substitute in (1), we get

$\displaystyle r= c-\left ( \dfrac{a.c}{a.b} \right )b$

Hence

$\displaystyle r= \dfrac{\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$

$\alpha \bar{a}+\beta \bar{b}+\gamma \bar{c}=0$ , then

$\left ( \bar{a}\times \bar{b} \right )\times \left [ \left ( \bar{b}\times \bar{c} \right )\times \left ( \bar{c}\times \bar{a} \right ) \right ]$ is equal to

0%
$\bar{0}$
0%
A vector $\perp$ plane of $\bar{a}$ , $\bar{b}$ and $\bar{c}$
0%
A scalar quantity
0%
None of these

Explanation

Since

$\alpha \bar{a}+\beta \bar{b}+\gamma \bar{c}=0$

$\therefore$ the vectors

$\bar{a}$ ,

$\bar{b}$ and

$\bar{c}$ are coplanar
and thus

$\left ( \bar{c}\times \bar{b} \right )$ and

$\left ( \bar{c}\times \bar{a} \right )$ are coplanar.

$\therefore$

$\left ( \bar{b}\times \bar{c} \right )\times \left ( \bar{c}\times \bar{a} \right )=0$
Hence option

$A$ becomes correct.

A unit vector perpendicular to each of the vectors

$\displaystyle 2\hat{i}+4\hat{j}-\hat{k}$ and

$\displaystyle \hat{i}-2\hat{j}+3\hat{k}$ forming a right handed system is

0%
$\displaystyle 7\hat{i}-10\hat{j}+8\hat{k}$
0%
$\displaystyle \frac{10\hat{i}-7\hat{j}-8\hat{k}}{\sqrt{213}}$
0%
$\displaystyle -7\hat{i}+10\hat{j}+8\hat{k}$
0%
$\displaystyle \frac{-10\hat{i}-7\hat{j}-8\hat{k}}{\sqrt{213}}$

Explanation

Unit vector perpendicular to the given 2 vectors is

$(2\hat{i} + 4\hat{j} - \hat{k}) \times (\hat{i} - 2\hat{j} + 3\hat{k})$

$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -1 \\ 1 & -2 & 3 \end{vmatrix}$

$= \hat{i}(12 - 2) - \hat{j}(6 + 1) + \hat{k}(- 4 - 4)$

$= 10 \hat{i} - 7\hat{j} - 8\hat{k}$
The unit vector will be obtained by dividing the obtained vector with its modulus.
So, we get the answer as

$\dfrac{10 \hat{i} - 7\hat{j} - 8\hat{k}}{\sqrt{100 + 49 + 64}} = \dfrac{10 \hat{i} - 7\hat{j} - 8\hat{k}}{\sqrt{213}}$

The position vector of the points $A$ and $B$ are respectively

$\bar{a}$ and

$\bar{b}$ divides

$AB$ in the ratio

$3:1$ and

$Q$ iis the midpoint of

$AP$ . The position vector of

$Q$ is

0%
$\displaystyle \dfrac{3\bar{a}+5\bar{b}}{8}$
0%
$\displaystyle\dfrac{3\bar{a}+\bar{b}}{4}$
0%
$\displaystyle\dfrac{\bar{a}+3\bar{b}}{4}$
0%
$\displaystyle\dfrac{5\bar{a}+3\bar{b}}{8}$

Explanation

Given

position vector of A

$\vec{OA}=\vec{a}$

position vector of B

$\vec{OB}=\vec{b}$

Point P divides AB in ration

$3:1$

position vector of Point P

$\vec{OP}=\dfrac { 3\bar { OB } +\bar { OA } }{ 4 }$

$\vec{OP}=\dfrac { 3\bar { b } +\bar { a} }{ 4 }$

Here Q is the mid point of AP

$\vec{OQ}=\dfrac{\vec{OA}+\vec{OP}}{2}$

$Q=\dfrac { \dfrac { 3\bar { b } +\bar { a } }{ 4 } +\bar { a } }{ 2 } =\dfrac { 5\bar { a } +3\bar { b } }{ 8 }$

For three unit vectors

$\bar{a}$ ,

$\bar{b}$ and

$\bar{c}$ , if

$\bar{a}+\bar{b}+\bar{c}= \bar{0}$ , then the value of

$\bar{a}\cdot \left ( \bar{b}+\bar{c} \right )+\bar{b}\cdot \left ( \bar{c}+\bar{a} \right )+\bar{c}\cdot \left ( \bar{a}+\bar{b} \right )$ is equal to

0%
$-\dfrac{3}{2}$
0%
$0$
0%
$-3$
0%
None of these

Explanation

Given

$\bar{a} + \bar{b} + \bar{c} = 0$
We have

$\therefore\bar{a}\cdot \left ( \bar{b}+\bar{c} \right )+\bar{b}\cdot \left ( \bar{c}+\bar{a} \right )+\bar{c}\cdot \left ( \bar{a}+\bar{b} \right )$

$= \bar{a}. - \bar{a} - \bar{b}.\bar{b} - \bar{c}.\bar{c}$

$= -a^2 - b^2 - c^2$
If all the three vectors are unit vectors, the answer will be

$-3$

$\displaystyle \bar{a}\times \bar{b}=\bar{b}\times \bar{c}=\bar{c}\times \bar{a}$ then

$\displaystyle \bar{a}+\bar{b}+\bar{c}=?$

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

Explanation

Let us consider

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

Multiply with

$\overline { b }$ on both sides , we get

$\overline { a }\times \overline { b } +\overline { b } \times\overline { b } +\overline { c }\times \overline { b } =k\times\overline { b }$

$\Rightarrow k \times \overline { b } =\overline { b } \times \overline { c }+0+\overline { c }\times \overline { b } =0$

Similarly we get

$k \times \overline { a } =0$ and

$k \times \overline { c } =0$

therefore

$k=0$ for equations to satisfy

therefore correct option is

$C$

The vector

$\left ( \bar{a }\times\bar{b } \right )\times \left ( \bar{c }\times\bar{b } \right )$ is

0%
At right angle to $\bar{b}$
0%
Is parallel to $\bar{c}$
0%
Is parallel to $\bar{b}$
0%
None of these

Explanation

$(a\times b)$ will be a vector perpendicular to

$\vec b$ .

$(c\times b)$ will be a vector perpendicular to

$\vec b$ .
Hence

$(a\times b)\times (c\times b)$ will be a vector perpendicular to both

$(a\times b)$ and

$(c\times b)$ .
Thus it will be a vector parallel to

$\vec b$ .

In a trapezium the vector

$\overline{BC} = \alpha \overline{AD}$ . We will then find that

$\bar{p}= \overline{AC}+\overline{BD}$ is collinear with

$\overline{AD}$ . if

$\bar{p}= \mu \overline{AD}$ then

0%
$\mu =\alpha +2$
0%
$\mu +\alpha = 2$
0%
$\alpha = \mu +1$
0%
$\mu = \alpha+1$

Explanation

Let

$A$ be the origin.
So,

$\alpha \vec{d} = \vec{c} - \vec{b}$

$p = \vec{c} + \vec{d} - \vec{b} = \mu{\vec{d}}$

Combining two equations, we get

$(1 + \alpha)\vec{d} = \mu \vec{d}$

$\Rightarrow \mu = \alpha + 1$

$C$ is the mid point of

$AB$ and

$P$ is any point outside

$AB$ , then

0%
$\displaystyle \overline{PA}+\overline{PB}+\overline{PC}=0$
0%
$\displaystyle \overline{PA}+\overline{PB}+2\overline{PC}=\bar{0}$
0%
$\displaystyle \overline{PA}+\overline{PB}=\overline{PC}$
0%
$\displaystyle \overline{PA}+\overline{PB}=2\overline{PC}$

Explanation

Let

$\vec{p}, \vec{a}, \vec{b}, \vec{c}$ be the position vectors of points

$P, A, B, C$ respectively.

$L.H.S = \overline{PA} + \overline{PB} = \vec{p} - \vec{a} + \vec{p} - \vec{b} = 2\left(\vec{p} - \frac{\vec{a} + \vec{b}}{2}\right) = 2\vec{PC}$

Given

$A=ai+bj+ck, \ B=di+3j+4k$ and

$C=3i+j-2k$ . If the vectors

$A,B$ and

$C$ form a triangle such that

$A=B+C$ and

$area(\Delta ABC)=5\sqrt6$ , then

0%
$a=-8, \ b=-4, \ c=2, \ d=-11$
0%
$a=-8, \ b=4, \ c=-2, \ d=-11$
0%
$a=-8,\ b=4, \ c=2, \ d=-11$
0%
None of the above

Explanation

Here

$A,B,C$ are the vectors which represent the sides of the triangle

$ABC$ where

$A=ai+bj+ck\\ B=di+3j+4k\\ C=3i+j-2k$

Given that

$A=B+C$

$\displaystyle \therefore ai+bj+ck=\left( d+3 \right) i+4j+2k\\ \Rightarrow a=d+3,b=4,c=2\\ B\times C=\begin{vmatrix} i & j & k \\ d & 3 & 4 \\ 3 & 1 & -2 \end{vmatrix}=-10i+\left( 2d+12 \right) j+\left( d-9 \right) k$

$\therefore$ Area of the

$\displaystyle \triangle ABC=\frac { 1 }{ 2 } \left| B\times C \right| =\frac { 1 }{ 2 } \sqrt { \left[ 100+{ \left( 2d+12 \right) }^{ 2 }+{ \left( d-9 \right) }^{ 2 } \right] } =5\sqrt { 6 }$

$\Rightarrow \sqrt { 5{ d }^{ 2 }+30d+325 } =10\sqrt { 6 } \Rightarrow 5{ d }^{ 2 }+30d+325=600\\ \Rightarrow 5{ d }^{ 2 }+30d-275=0\Rightarrow { d }^{ 2 }+6d-55=0\\ \Rightarrow \left( d+11 \right) \left( d-5 \right) =0\Rightarrow d=5,-11$

When

$d=5,a=8,b=4,c=2$ and when

$d=-11,a=-8,b=4,c=2$

Two identical particles are located at

$\overrightarrow{x}$ and

$\overrightarrow{y}$ with reference to the origin of three dimensional co-ordinate system. The position vector of centre of mass of the system is given by

0%
$\overrightarrow { x } - \overrightarrow { y }$
0%
$\displaystyle \frac {\overrightarrow { x } + \overrightarrow { y }} {{2}}$
0%
$(\overrightarrow { x } - \overrightarrow { y })$
0%
$\displaystyle \frac{\overrightarrow { x } - \overrightarrow { y }} {{2}}$

A vector

$a$ can be written as

0%
$(a\cdot i)i+(a\cdot j)j+(a\cdot k)k$
0%
$(a\cdot j)i+(a\cdot k)j+(a\cdot i)k$
0%
$(a\cdot k)j+(a\cdot i)j+(a\cdot j)k$
0%
$(a\cdot a)i+(i+j+k) k$

Explanation

Let

$a=a_{1} i+a_{2} i+a_{3} k$ , so that

$a.i=a _{1},a\cdot j=a_{2}$ and

$a.k=a_{3}$

$\Rightarrow a=(a\cdot i)i+(a\cdot j)j+(a\cdot k)k$

$b \neq 0,$ then every vector

$a$ can be written in a unique manner as the sum of a vector

$a_{||}$ parallel to b and a vector

$a_{\perp}$ perpendicular to

$b$ . If

$a$ is parallel to

$b,$ then

$a_{||} = a$ and

$a_{\perp } =0$ . If

$a$ is perpendicular to

$b$ , then

$a_{||} = 0$ and

$a_{\perp} = a$ . The vector

$a_{||}$ is called the projection of

$a$ on

$b$ and is denoted by

$proj_{b}a$ . Since

$proj_{b} a$ is parallel to

$b$ , it is a scalar multiple of the unit vector in the direction of

$b$ , i.e.,

$proj _{b} a =\lambda u_{b}$
The scalar

$\lambda$ is called the component of

$a$ in the direction of

$b$ and is denoted by

$comp _{b} a$ . In fact,

$proj _{b} a = (a \cdot u_{b})u_{b}$ and

$comp _{b} a = a \cdot u_{b}$

$a = -2i + j + k$ and

$b =4i -3j + k$ , then

$proj_{b} a$ is equal to

0%
$4i -3j + 2k$
0%
$\displaystyle -\frac{ 5}{13} (4i- 3j + k)$
0%
$\displaystyle \frac{ 5}{13} (4i -3j+ k)$
0%
$\displaystyle -\frac{ 4}{11} (2i -j + 2k)$

Explanation

$\displaystyle \left| b \right| =\sqrt { 26 } ,{ u }_{ b }=\frac { b }{ \left| b \right| } =\frac { 1 }{ \sqrt { 26 } } \left( 4i-3j+k \right)$

$\displaystyle { comp }_{ b }a=a.{ u }_{ b }=\frac { -10}{ \sqrt{26} }$

$\displaystyle { Proj }_{ b }a=\left( { comp }_{ b }a \right) { u }_{ b }=-\frac { 5 }{ 13 } \left( 4i-3j+k \right)$

$ABCD$ is a quadrilateral and

$E$ the point of intersection of the lines joining the middle points of opposite sides. If

$O$ is any point, then the resultant of

$OA,OB,OC$ and

$OD$ is equal to

0%
$2OE$
0%
$OE$
0%
$4OE$
0%
none of these

Explanation

Let

$P,Q,R,S$ be the mid-points of sides

$BC,CD,DA,AB$ respectively of a quadrilateral

$ABCD$ .

BY geometry the figure formed by joining the mid-point

$P,Q,R,S$ will be a parallelogram.

Hence, its diagonal will bisect each other, say at

$E$ .

Now

$\because P$ is the mid point of

$BC$

$\therefore OB+OC=2OP$ ....(1)

And

$R$ is the mid point of

$AD$

$\therefore OA+OD=2OR$ ....(2)

Adding (1) and (2), we have

$OA+OB+OC+OD=2(OP+OR)=2.2OE=4OE$

$(\because E$ is mid point of

$PR$ ,

$\therefore OP+OR=2OE)$ .

The components of vector

$i + j + k$ along vector

$i+2j+3k$ is

0%
$(3/7) (i + 2j + 3k)$
0%
$(i + 2j + 3k)$
0%
$(1/7) (i + 2j + 3k)$
0%
$(4/7) (i + 2j + 3k)$

Explanation

The projection of vector

$i + j + k$ along vector,

$i+2j+3k$ is

$\quad =\displaystyle \frac{(i+j+k)\cdot (i+2j+3k)}{|i+2j+3k|}=\frac{6}{\sqrt{14}}$
Hence component of vector

$i+j+k$ along

$i+2j+3k$ is,

$\quad =\displaystyle \frac{6}{\sqrt{14}}.\frac{i+2j+3k}{|1+2j+3k|}=\frac{3}{7}(i+2j+3k)$

Four forces of magnitude P, 2P, 3P and 4P act along the four sides of a square ABCD in cyclic order. Use the vector method to find the resultant force.

0%
$\displaystyle 2\sqrt{2}P$
0%
$\displaystyle 3\sqrt{2}P$
0%
$\displaystyle \sqrt{2}P$
0%
$\displaystyle -2\sqrt{2}P$

Explanation

The resultant of 4 vectors in X, Y direction are:

resultant in X direction is

$3\vec P - \vec P = 2\vec P$ (in -ve x direction)

In y direction is

$4\vec P - 2\vec P = 2\vec P$ (in +ve y direction)

So, the net resultant force vector is

$\overset { \rightarrow }{ R } =2\overset { \rightarrow }{ P } +2\overset { \rightarrow }{ P }$

angle between both vectors is 90

So magnitude of resultant is

$2\sqrt { 2 } P$ in direction of

${ 45 }^{ 0 }$ from positive y axis in IV quadrant.

OAB is a given triangle such that

$\displaystyle \overrightarrow{OA}=\bar a$ ,

$\displaystyle \overrightarrow{OB}=\bar b$ .
Also C is a point on AB such that

$\overrightarrow{AB}=2\overrightarrow{BC}$ . State which of the following statements are correct?

0%
$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left ( \bar b-\bar a \right )$ ,
0%
$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left ( \bar a-\bar b \right )$ ,
0%
$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left (\bar b+\bar a \right )$ ,
0%
$\displaystyle \overrightarrow{AC}=\frac{3}{2}\left ( \bar b-\bar a \right )$ ,

Explanation

Given

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

$\overline{OB}=\bar b$

$C$ is apoint on

$AB$ such that

$\overline{AB}=2\overline{BC}$

$\Rightarrow \bar b - \bar a = 2(\bar c - \bar b)$

$\Rightarrow \bar c=\displaystyle\dfrac{3\bar b-\bar a}{2}$

$\therefore \overline{AC}=\bar c-\bar a =\displaystyle\dfrac{3}{2}(\bar b-\bar a)$
Hence, option D.

$ABC$ is any triangle and

$D, E, F$ are the middle points of its sides

$BC, CA, AB$ respectively. Express the vectors

$\displaystyle \overrightarrow{CF}$ and

$\overrightarrow{BE}$ as linear combination of the vectors

$\displaystyle \overrightarrow{AB}$ and

$\overrightarrow{AC}$

0%
$\displaystyle \overrightarrow{BE}= \frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB};\,\,\,\overrightarrow{CF}= \frac{1}{2}\overrightarrow{AB}-\overrightarrow{AC}$
0%
$\displaystyle \overrightarrow{BE}= \frac{1}{2}\overrightarrow{AC}+\overrightarrow{AB};\,\,\,\overrightarrow{CF}= \frac{1}{2}\overrightarrow{AB}+\overrightarrow{AC}$
0%
$\displaystyle \overrightarrow{CF}= \frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB};\,\,\,\overrightarrow{BE}= \frac{1}{2}\overrightarrow{AB}-\overrightarrow{AC}$
0%
$\displaystyle \overrightarrow{CF}= \frac{1}{2}\overrightarrow{AC}+\overrightarrow{AB};\,\,\,\overrightarrow{BE}= \frac{1}{2}\overrightarrow{AB}+\overrightarrow{AC}$

Explanation

$\overrightarrow { CA } +\overrightarrow { AF } =\overrightarrow { CF }$

$\Rightarrow -\overrightarrow { AC } +\dfrac { 1 }{ 2 } \overrightarrow { AB } =\overrightarrow { CF }$
Similarly

$\overrightarrow { AE } +\overrightarrow { BA } =\overrightarrow { BE }$

$\Rightarrow \dfrac { 1 }{ 2 } \overrightarrow { AC } -\overrightarrow { AB } =\overrightarrow { BE }$

$A\left( \overrightarrow { a } \right) ,B( \overrightarrow { b } ) ,C\left( \overrightarrow { c } \right)$ be the vertices of a triangle whose circumcentre is the origin, then orthocenter is given by

0%
$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 }$
0%
$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 2 }$
0%
$\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }$
0%
None of these

Explanation

$A\left( \overrightarrow { a } \right) ,B\left( \overrightarrow { b } \right) ,C\left( \overrightarrow { c } \right)$ ate the vertices of the

$\triangle ABC$ .

Circumcentre is at origin

$\displaystyle G\left( \overrightarrow { g } \right) =\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 }$

$\displaystyle G=\dfrac { 2\overrightarrow { c } +\overrightarrow { O } }{ 3 } \Rightarrow \overrightarrow { O } =3\overrightarrow { G }$

$\displaystyle \Rightarrow \overrightarrow { O } =3\left( \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } \right) =\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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