If $$a$$ is perpendicular to $$b$$ and $$c$$, then

$$\displaystyle a\times \left ( b\times c \right )=0$$

$$\displaystyle a\times \left ( b\times c \right )=1$$

$$\displaystyle a\times \left ( b\times c \right )=-1$$

MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 6

The projection of

$\bar a=3\hat i-\hat j +5\hat k$ on

$\bar b=2\hat i+3\hat j+\hat k$ is

0%
$\displaystyle 8/\sqrt{(35)}$
0%
$\displaystyle 8/\sqrt{(39)}$
0%
$\displaystyle 8/\sqrt{(14)}$
0%
$\displaystyle \sqrt{(14)}$

Explanation

The projection of

$\bar a=3\hat i-\hat j +5\hat k$ on

$\bar b=2\hat i+3\hat j+\hat k$ is

$\displaystyle\frac{\bar a\cdot \bar b}{|\bar b|}$

$=\displaystyle\frac{(3\hat i-\hat j +5\hat k)\cdot (2\hat i+3\hat j+\hat k) }{|2\hat i+3\hat j+\hat k|}$

$=\displaystyle\frac{8}{\sqrt {14}}$
Hence, option C.

The triangle

$ABC$ is defined by the vertices

$A (1, -2, 2), B(1, 4, 0)$ and

$C(-4, 1, 1)$ . Let

$M$ be the foot of the altitude drawn from the vertices

$B$ to side

$AC$ . Then

$\displaystyle \overrightarrow{BM}=$

0%
$\left (-\dfrac {20}{7}, -\dfrac {30}{7}, \dfrac {10}{7}\right)$
0%
$(-20, -30, 10)$
0%
$(2, 3, -1)$
0%
none of these

Explanation

$\displaystyle\overrightarrow { MB } =\overrightarrow { AB } -\overrightarrow { AM } =\overrightarrow { AB } -\dfrac { \left( \overrightarrow { AB } .\overrightarrow { AC } \right) \overrightarrow { AC } }{ { \left( \overrightarrow {| AC |} \right) }^{ 2 } }$

$\displaystyle=\left( 6\vec j-2\vec k \right) -\dfrac { \left( 6\vec j-2\vec k \right) .\left( 5\vec i+3\vec j-\vec k \right) \left( -5\vec i+3\vec j-\vec k \right) }{ \left( 25+9+1 \right) }$

$\displaystyle=\left( 6\vec j-2\vec k \right) -\dfrac { 20 }{ 35 } \left( -5\vec i+3\vec j-\vec k \right)$

$\displaystyle=\dfrac { 10 }{ 7 } \left( 2\vec i+3\vec j-k \right)$

$\displaystyle\therefore \overrightarrow { BM } =-\dfrac { 10 }{ 7 } \left( 2\vec i+3\vec j-\vec k \right) =\left( -\dfrac { 20 }{ 7 } \vec i-\dfrac { 30 }{ 7 } \vec j+\dfrac { 10 }{ 7 }\vec k \right)$

Given two vectors

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

$b=-2\hat{i}+2\hat{j}-1\hat{k}$ and

$\displaystyle \lambda$ = ratio of the projection of

$a$ on

$b$ and the projection of

$b$ on

$a,$ then the value of

$\displaystyle \lambda$ is

0%
$\dfrac{3}{7}$
0%
$\dfrac{7}{3}$
0%
$3$
0%
$7$

Explanation

Projection of

$\vec{b}$ on

$\vec{a}$ will be

$=bcos\theta$

$=\dfrac{\vec{a}.\vec{b}}{a}$ ...(i)

Similarly

Projection of

$\vec{a}$ on

$\vec{b}$ will be

$=acos\theta$

$=\dfrac{\vec{a}.\vec{b}}{b}$ ...(ii)

Therefore

$\lambda=\dfrac{ii}{i}$

$=\dfrac{\dfrac{\vec{a}.\vec{b}}{b}}{\dfrac{\vec{a}.\vec{b}}{a}}$

$=\dfrac{a}{b}$

$=\dfrac{\sqrt{2^{2}+3^{2}+6^{2}}}{\sqrt{2^{2}+2^{2}+1^{2}}}$

$=\dfrac{\sqrt{49}}{\sqrt{9}}$

$=\dfrac{7}{3}$

If c is the middle point of AB and P is any point outside AB then

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

Explanation

$\overrightarrow { AC }$ and

$\overrightarrow { BC }$
have equal magnitude and opposite direction
Therefore both are negative vectors of each other

$\therefore \overrightarrow { AC } +\overrightarrow { BC } =0$
In

$\triangle PAC$ we have

$\overrightarrow { PA } +\overrightarrow { AC } =\overrightarrow { PC }$ ...(1)
In

$\triangle PBC$ , we have

$\overrightarrow { PB } +\overrightarrow { BC } =\overrightarrow { PC }$ ...(2)
From (1) and (2)

$\overrightarrow { PA } +\overrightarrow { PB } +\left( \overrightarrow { AC } +\overrightarrow { BC } \right) =2\overrightarrow { PC } \\ \Rightarrow \overrightarrow { PA } +\overrightarrow { PB } =2\overrightarrow { PC }$

Let

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

$c=i+j-k$ . A vector in the plane of

$a$ and

$b$ , whose projection on

$c$ is

$\displaystyle \frac{1}{\sqrt{3}}$ is

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

Explanation

A vector

$d$ in the plane of

$a$ and

$b$ is given by,

$d =\lambda a+\mu b =(\lambda +\mu)i+(2\lambda -\mu)j+(\lambda+\mu)k,$
Now the projection of

$d$ on

$c$ is

$\displaystyle \frac{c\cdot d}{\left|c\right|}=\frac{\lambda +\mu+(2\lambda-\mu)-(\lambda+\mu)}{\sqrt{3}}=\frac{1}{\sqrt{3}}$ (given)

$\Rightarrow 2\lambda -\mu=1$ . Thus

$d=(3\lambda-1)i+j+(3\lambda-1)k,$
So for

$\lambda =-1$ ,

$d=- 4i+j-4k$

ABCD is a parallelogram and AC, BD are its diagonals Express

$\displaystyle \overrightarrow{AC}\:and\:\overrightarrow{BD}$ in terms of

$\displaystyle \overrightarrow{AB}\:and\:\overrightarrow{AD}$ only

0%
$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{AD}\quad;\:\overrightarrow{BD}= -\overrightarrow{AB}+\overrightarrow{AD}$
0%
$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}-\overrightarrow{CB}\quad;\:\overrightarrow{BD}= -\overrightarrow{AB}+\overrightarrow{AD}$
0%
$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{AD}\quad;\:\overrightarrow{BD}= \overrightarrow{BC}+\overrightarrow{CD}$
0%
all of the above

Explanation

Given

$ABCD$ is a parallelogram and

$AC, BD$ are its diagonals.

$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{BC};\:\overrightarrow{BD}= \overrightarrow{BA}+\overrightarrow{AD}$

$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{AD};\:\overrightarrow{BD}= -\overrightarrow{AB}+\overrightarrow{AD}$
Hence, option A.

Given a cube

$ABCD{ A }_{ 1 }{ B }_{ 1 }{ C }_{ 1 }{ D }_{ 1 }$ with lower base

$ABCD$ , upper base

${ A }_{ 1 }{ B }_{ 1 }{ C }_{ 1 }{ D }_{ 1 }$ and the lateral edges

$A{ A }_{ 1 },B{ B }_{ 1 },C{ C }_{ 1 }$ and

$D{ D }_{ 1 }$ ;

$M$ and

${ M }_{ 1 }$ are the centers of the faces

$ABCD$ and

${ A }_{ 1 }{ B }_{ 1 }{ C }_{ 1 }{ D }_{ 1 }$ respectively.

$O$ is apoint on line

$M{ M }_{ 1 }$ , such that

$OA+OB+OC+OD=O{ M }_{ 1 }$ , then

$OM=\lambda O{ M }_{ 1 }$ is

$\lambda=$

0%
$\displaystyle \frac { 1 }{ 4 }$
0%
$\displaystyle \frac { 1 }{ 2 }$
0%
$\displaystyle \frac { 1 }{ 6 }$
0%
$\displaystyle \frac { 1 }{ 8 }$

Explanation

$OM_1=OA+OB+OC+OD$ (given)

$=OM+MA+OM+MB+OM+MC+OM+MD$

$=4OM+(MA+MC)+(MB+MD)=4OM$

$(\because MA=-MC,MB=-MD)$

$\displaystyle\therefore OM=\frac{1}{4}OM_1,$

$\displaystyle \therefore \lambda=\frac{1}{4}$

$P, Q, R, S$ have position vectors

$p, q, r, s$ , such that

$\displaystyle p-q=2\left ( s-r \right )$ , then

0%
$PQ$ and $RS$ bisect each other
0%
$PQ$ and $PR$ bisect each other
0%
$PQ$ and $RS$ trisect each other
0%
$QS$ and $PR$ trisect each other

Explanation

$\overrightarrow { p } -\overrightarrow { q } =2\left( \overrightarrow { s } -\overrightarrow { r } \right) \Rightarrow \overrightarrow { p } +2\overrightarrow { r } =2\overrightarrow { s } +\overrightarrow { q } =\overrightarrow { q } +2\overrightarrow { s }$

$\displaystyle\Rightarrow \dfrac { \overrightarrow { p } +\overrightarrow { 2 } r }{ 1+2 } =\dfrac { \overrightarrow { q } +2\overrightarrow { s } }{ 1+2 }$

$\therefore$ the point divides

$PR$ in the ratio

$2:1$ , divides

$QS$ in the ratio

$2:1$ .

$\therefore$

$PR$ and

$QS$ trisect each-other

Which of the following statements are correct:
If

$M$ is the mid-point of

$AB$ and

$O$ is any point then

0%
$\displaystyle \overrightarrow{OM}= \overrightarrow{OA}+\overrightarrow{MA}$
0%
$\displaystyle \overrightarrow{OM}= \overrightarrow{OA}-\overrightarrow{MA}$
0%
$\displaystyle \overrightarrow{OM}= \dfrac{1}{2}\left ( \overrightarrow{OA}-\overrightarrow{OB} \right )$
0%
$\displaystyle \overrightarrow{OM}= \dfrac{1}{2}\left ( \overrightarrow{OB}+\overrightarrow{OA} \right )$

Explanation

Given, M is the mid-point of AB.

$\displaystyle\overrightarrow{OM}= \overrightarrow{OA}+\overrightarrow{AM}=\overrightarrow{OA}-\overrightarrow{MA}$

$\displaystyle\overrightarrow{AM}=\overrightarrow{MB}$

$\Rightarrow\displaystyle\overrightarrow{OM}-\overrightarrow{OA}=\overrightarrow{OB}-\overrightarrow{OM}$

$\Rightarrow\displaystyle\overrightarrow{OM}=\dfrac{\overrightarrow{OA}+\overrightarrow{OB}}{2}$
Hence, options B and D.

OABC is a tetrahedron express the vectors

$\displaystyle \overrightarrow{BC},\overrightarrow{CA},\:and\:\overrightarrow{AB}$ in terms of the vectors

$\displaystyle \overrightarrow{OA},\overrightarrow{OB}\:and\:\overrightarrow{OC}$

0%
$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}+\overrightarrow{OB}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}+\overrightarrow{OC}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}+\overrightarrow{OA}$
0%
$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}-\overrightarrow{OB}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}-\overrightarrow{OC}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}-\overrightarrow{OA}$
0%
$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}-\overrightarrow{BO}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}-\overrightarrow{CO}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}-\overrightarrow{AO}$
0%
none of these

Explanation

$OABC$ is a tetrahedron

$\displaystyle \overrightarrow{BC}=\overrightarrow{BO}+\overrightarrow{OC}=\overrightarrow{OC}-\overrightarrow{OB}$

$\displaystyle\overrightarrow{CA}=\overrightarrow{CO}+\overrightarrow{OA}= \overrightarrow{OA}-\overrightarrow{OC}$

$\displaystyle\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}= \overrightarrow{OB}-\overrightarrow{OA}$
Hence, option B.

Given the vectors

$\bar a$ and

$\bar b$ as follows. Find the projections of

$\bar a$ on

$\bar b$ and of

$\bar b$ on

$\bar a$ .

$\bar a=\hat i+\hat j+\hat k$ ;

$\displaystyle \bar b= \sqrt{3}\hat i+3\hat j-2\hat k$

0%
$\displaystyle \frac{\sqrt{2}+1}{5},\frac{3+\sqrt{3}}{5}.$
0%
$\displaystyle \frac{\sqrt{3}+1}{4},\frac{3+\sqrt{3}}{3}.$
0%
$\displaystyle \frac{\sqrt{5}+2}{4},\frac{4+\sqrt{3}}{4}.$
0%
$\displaystyle \frac{\sqrt{2}+5}{3},\frac{4+\sqrt{3}}{3}.$

Explanation

Given

$\bar a=\hat i+\hat j+\hat k$ ;

$\displaystyle \bar b= \sqrt{3}\hat i+3\hat j-2\hat k$
Projection of

$\bar a$ on

$\bar b$ is

$\displaystyle\frac{\bar a \cdot \bar b}{|\bar b|}=\frac{\sqrt{3}+1}{4}$
Projection of

$\bar b$ on

$\bar a$ is

$\displaystyle\frac{\bar a \cdot \bar b}{|\bar a|}=\frac{\sqrt{3}+1}{\sqrt{3}}=\frac{3+\sqrt{3}}{3}$
Hence, option

$B$ .

$\displaystyle \overrightarrow{PO}+\overrightarrow{OQ}=\overrightarrow{QO}+\overrightarrow{OR},$ then

$\displaystyle P, Q, R$ are

0%
the vertices of an equilateral triangle
0%
the vertices of an isosceles triangle
0%
collinear
0%
None of these

Explanation

$\displaystyle \overrightarrow{PO}+\overrightarrow{OQ}=\overrightarrow{QO}+\overrightarrow{OR}$

$\Rightarrow\displaystyle 2\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{OR}$

$\Rightarrow \displaystyle\overrightarrow{OQ}=\dfrac{\overrightarrow{OP}+\overrightarrow{OR}}{2}$

$\therefore P,Q,R$ are collinear.
Hence, option C.

Which of the following statements are correct:
If in triangle OAC, B is the mid-point of AC and

$\displaystyle \overrightarrow{OA}= \vec{a}\:$ and

$\:\overrightarrow{OB}= \vec{b}$ then

0%
$\displaystyle \overrightarrow{OC}= \frac{1}{2}\left ( \vec{a}+\vec{b} \right )$
0%
$\displaystyle \overrightarrow{OC}= 2\vec{b}-2\vec{a}$
0%
$\displaystyle \overrightarrow{OC}= 2\vec{b}-\vec{a}$
0%
$\displaystyle \overrightarrow{OC}= 3\vec{a}-2\vec{b}$

Explanation

$B$ is the midpoint of

$AC$

$\Rightarrow \overrightarrow { OB } =\dfrac { \overrightarrow { OA } +\overrightarrow { OC } }{ 2 }$

$\therefore \overrightarrow { OC } =2\overrightarrow { OB } -\overrightarrow { OA } =2\overrightarrow { b } -\overrightarrow { a }$

Hence, option C.

If a, b, c are position vectors of the vertices of a

$\displaystyle \Delta ABC,$ then

$\displaystyle \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=$

0%
0
0%
2a
0%
2b
0%
3c

Projection of the vector

$2i + 3j - 2k$ on the vector

$i - 2j + 3k$ is

0%
$\displaystyle 2/\sqrt{\left ( 14 \right )}$
0%
$\displaystyle 1/\sqrt{\left ( 14 \right )}$
0%
$\displaystyle 3/\sqrt{\left ( 14 \right )}$
0%
None of these

Explanation

Projection of

$a$ and

$b$ is

$\displaystyle \dfrac { a\cdot b }{ \left| b \right| } =\dfrac { \left( 2i+3j-2k \right) \cdot \left( i-2j+3k \right) }{ \left| i-2j+3k \right| } =\dfrac { 2 }{ \sqrt { 14 } }$

Given that the vectors

$\bar a,\bar b,\bar c$ form a base, find the sum of co-ordinates of the vector:

$3\bar u-\bar v+\bar w$

$\bar u=\bar a+\bar c, \bar v=\bar b+\bar c,\bar w=\bar a-\bar b$ ;

0%
4
0%
6
0%
2
0%
3

Explanation

Given vectors

$\bar{u} = \bar{a} + \bar{c}$

$\bar{v} = \bar{b} + \bar{c}$

$w = \bar{a} - \bar{b}$

$3 \bar{u} - \bar{v} + \bar{w} = 3(a + c) - (b + c) + (a - b)$

$= 3a + 3c - b - c + a - b$

$= 4a - 2b + 2c$ .

Sum of the co-ordinates of the vector

$= 4 + (-2) + 2 = 4 - 2 + 2 = 4$ .

$\therefore$ The given option is correct.

$\displaystyle \left ( \bar A+\bar B \right )$ is perpendicular to

$\bar B$ and If

$\displaystyle \left ( \bar A+2\bar B \right )$ is perpendicular to

$\bar A$ , then

0%
$\displaystyle A=\sqrt{2}B$
0%
$\displaystyle A=2B$
0%
$\displaystyle 2A=B$
0%
$\displaystyle A=B$

Explanation

Given

$\displaystyle \left ( \bar A+\bar B \right )$ is perpendicular to

$\bar B$

$\Rightarrow \displaystyle \left ( \bar A+\bar B \right )\cdot \bar B=0$

$\Rightarrow \bar A\cdot \bar B + B^2=0$ -----(1)
And

$\displaystyle \left ( \bar A+2\bar B \right )$ is perpendicular to

$\bar A$ .

$\Rightarrow ( \bar A+2\bar B )\cdot \bar A=0$

$\Rightarrow A^2+2\bar A\cdot \bar B=0$ ------(2)
from (1) and (2)

$A^2=2B^2$

$\therefore A=\sqrt{2}B$
Hence, option A.

$\displaystyle \overrightarrow{AE}$

0%
$\displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar a+\bar b \right )$
0%
$\displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar a-\bar b \right )$
0%
$\displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar b-\bar a \right )$
0%
none of these

Explanation

Given

$ABCD$ is a parallelogram in which

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

$\overline{AD}=\bar b$ .
And

$E$ is point of intersection of diagonals.

$\overline{AE}=\displaystyle\frac{1}{2}\overline{AC}$ As, diagonals bisect each other.

$=\displaystyle\frac{1}{2}(\bar a+\bar b)$

$\therefore \displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar a+\bar b \right )$
Hence, option A.

$a$ is perpendicular to

$b$ and

$c$ , then

0%
$\displaystyle a\times \left ( b\times c \right )=1$
0%
$\displaystyle a\times \left ( b\times c \right )=0$
0%
$\displaystyle a\times \left ( b\times c \right )=-1$
0%
None of these

Explanation

$\overrightarrow { a } \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) =\left( \overrightarrow { a } \cdot \overrightarrow { c } \right) \overrightarrow { b } -\left( \overrightarrow { a } \cdot \overrightarrow { b } \right) \overrightarrow { c } =0-0=0\\ \left( \because \overrightarrow { a } \bot \overrightarrow { b } ,\overrightarrow { a } \bot \overrightarrow { c } ,\therefore \overrightarrow { a } \cdot \overrightarrow { b } =0,\overrightarrow { a } \cdot \overrightarrow { c } =0 \right)$

$AB=3i+j-k$ and

$AC=i-j+3k$ . If the point

$P$ on the line segment

$BC$ is equidistant from

$AB$ and

$AC,$ then

$AP$ is

0%
$2i-k$
0%
$i-2k$
0%
$2i+k$
0%
None of these

Explanation

A point equidistant from

$AB$ and

$AC$ is on the bisector of the angle

$BAC.$

A vector along the internal bisector of the angle

$BAC$

$\displaystyle=\frac { AB }{ \left| AB \right| } +\frac { AC }{ \left| AC \right| }$

$\displaystyle=\frac { 3i+j-k }{ \sqrt { 9+1+1 } } +\frac { i-j+3k }{ \sqrt { 1+1+9 } } =\frac { 1 }{ \sqrt { 11 } } \left( 4i+2k \right) \\ \therefore AP=t\left( 2i+k \right) \\ \therefore BP=AP-AB=t\left( 2i+k \right) -\left( 3i+j-k \right) =\left( 2t-3 \right) i-j+\left( t+1 \right) k$

Also

$BC=AC-AB=\left( i-j+3k \right) -\left( 3i+j-k \right) =-2i-2j+4k$

But

$BP=sBC$

$\therefore \left( 2t-3 \right) i-j+\left( t+1 \right) k=s\left( -2i-2j+4k \right) \\ \therefore 2t-3=-2s,-1=-2s,t+1=4s$

$\displaystyle\therefore s=\frac { 1 }{ 2 }$ and

$t=1$

$\therefore AP=2i+k$

The points

$O,A,B,C$ are the vertices of a pyramid and

$P,Q,R,S$ are the mid-points of

$OA,OB,BC,AC$ respectively. If

$\displaystyle \overrightarrow{OA}=a,\overrightarrow{OB}=b,\overrightarrow{OC}=c,$ express in terms of

$a, b, c$ the vectors

$\displaystyle \overrightarrow{OP},\overrightarrow{OQ},\overrightarrow{OR}$ and

$\displaystyle \overrightarrow{OS}$ /

0%
$\displaystyle \overrightarrow{OP}=\dfrac{1}{2}a,$ $\overrightarrow{OQ}=\dfrac{1}{2}b,$ $\overrightarrow{OR}=\dfrac{1}{2}\left ( b+c \right ),\overrightarrow{OS}=\dfrac{1}{2}\left ( a+c \right )$
0%
$\displaystyle \overrightarrow{OP}=\frac{1}{2}c,\overrightarrow{OQ}=b,\overrightarrow{OR}=\frac{1}{2}\left ( b+c \right ),\overrightarrow{OS}=\frac{1}{2}\left ( a+c \right )$
0%
$\displaystyle \overrightarrow{OP}=\frac{1}{2}c,\overrightarrow{OQ}=b,\overrightarrow{OR}=\frac{1}{2}\left ( a+c \right ),\overrightarrow{OS}=\frac{1}{2}\left ( b+c \right )$
0%
$\displaystyle \overrightarrow{OP}=\frac{1}{2}a,\overrightarrow{OQ}=b,\overrightarrow{OR}=\frac{1}{2}\left ( a+c \right ),\overrightarrow{OS}=\frac{1}{2}\left ( b+c \right )$

Explanation

P,Q,R,S are the mid-points of OA,OB,BC,AC

$\overrightarrow { P } =\dfrac { \overrightarrow { A }-\vec{O} }{ 2 } =\dfrac { \overrightarrow { a } }{ 2 }$

$\vec{OP}=\vec{P}-\vec{O}$

$\vec{OP}=\dfrac{a}{2}-0$

$\vec{OP}=\dfrac{a}{2}$

$\overrightarrow { Q } =\dfrac { \overrightarrow { b } }{ 2 }$

$\vec{OQ}=\vec{Q}-\vec{O}$

$\vec{OP}=\dfrac{b}{2}-0$

$\vec{OP}=\dfrac{b}{2}$

$\overrightarrow { R } =\dfrac { \overrightarrow { b } +\overrightarrow { c } }{ 2 }$

$\vec{OR}=\vec{R}-\vec{O}$

$\vec{OR}=\dfrac{b+c}{2}-0$

$\vec{OR}=\dfrac{b+c}{2}$

$\overrightarrow { S } =\dfrac { \overrightarrow { a } +\overrightarrow { c } }{ 2 }$

$\vec{OS}=\vec{S}-\vec{O}$

$\vec{OS}=\dfrac{a+c}{2}-0$

$\vec{OS}=\dfrac{a+c}{2}$

$4i+7j+8k, 2i+7j+7k$ and

$3i+5j+7k$ are the position vectors of the vertices

$A,B$ and

$C$ respectively of triangle

$ABC$ . The position vector of the point where the bisector of angle

$A$ meets

$BC.$

0%
$\displaystyle \frac { 1 }{ 3 } \left( 5j+12k \right)$
0%
$\displaystyle \frac { 1 }{ 3 } \left( 6i+13j+18k \right)$
0%
$\displaystyle \frac { 2 }{ 3 } \left( 6i+8j+6k \right)$
0%
$\displaystyle \frac { 2 }{ 3 } \left( -6i-8j-6k \right)$

Explanation

Suppose the bisector of angle

$A$ meets

$BC$ at

$D$ .

Then,

$AD$ divides

$BC$ in the ratio

$AB:AC$

$\therefore$ position vector of

$D$

$\displaystyle =\frac { \left( AC \right) \left( 2i+3j+4k \right)+ \left( AB \right) \left( 2i+5j+7k \right) }{ \left( AB \right) +\left( AC \right) }$

But

$AB+-2i-4j-4k$ and

$AC=-2i-2j-k$

$\therefore AB= \sqrt{4+16+11}=6$ and

$AC=\sqrt{4+4+1}=3$

$\therefore$ P.V. of

$D$ is

$\displaystyle \frac { 6\left( 2i+5j+7k \right) +3\left( 2i+3j+4k \right) }{ 6+3 }$

$\displaystyle =\frac { 18i+39j+54k }{ 9 } =\frac { 1 }{ 3 } \left( 6i+13j+18k \right)$

Given two vectors

$\displaystyle a=2i-3j+6k$ on

$\displaystyle b=2i+3j-k$ and

$\displaystyle \lambda =\frac{the\:projection\:of\:a\:on\:b}{the\:projection\:of\:b\:on\:a},$ then the value of

$\displaystyle \lambda$ is

0%
3/7
0%
7/3
0%
3
0%
7

Explanation

$\displaystyle \lambda =\dfrac{the\:projection\:of\:a\:on\:b}{the\:projection\:of\:b\:on\:a},$

$\displaystyle =\dfrac { \dfrac { a.b }{ \left| b \right| } }{ \dfrac { b.a }{ \left| a \right| } } =\dfrac { \left| a \right| }{ \left| b \right| } =\dfrac { 7 }{ 3 }$

$\displaystyle \overrightarrow{BC}$

0%
$\displaystyle \overrightarrow{BC}=2(b-a)$
0%
$\displaystyle \overrightarrow{BC}=2(a-b)$
0%
$\displaystyle \overrightarrow{BC}=2(b+a)$
0%
none of these

Explanation

Given

$\vec{AB}=2\vec{a}$

$\vec{AC}=2\vec{b}$

by definition of position w.r.t origin O

$\vec{AB}=\vec{OB}-\vec{OA}$

$2\vec{a}=\vec{OB}-\vec{OA}$

$\vec{OB}=\vec{OA}+2\vec{a}$ ----------(1)

$\vec{AC}=\vec{OC}-\vec{OA}$

$2\vec{b}=\vec{OC}-\vec{OA}$

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

$\vec{BC}=\vec{OC}-\vec{OB}$

from eq (1) and (2)

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

$\vec{BC}=2\vec{b}+\vec{OA}-\vec{OA}-2\vec{a}$

$\vec{BC}=2\vec{b}-2\vec{a}$

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

The position vectors of three consecutive vertices of a parallelogram are

$i+j+k, i+3j+5k$ and

$7i+9j+11k$ . The position vector of the fourth vertex is

0%
$6(i+j+k)$
0%
$7(i+j+k)$
0%
$2j-4k$
0%
$6i+8j+10k$

Explanation

Let

$A\left( \overrightarrow { a } \right) =i+j+k$

$B\left( \overrightarrow { b } \right) =i+3j+5k\\ C\left( \overrightarrow { c } \right) =7i+9j+11k$

$D\left( \overrightarrow { d } \right)$ is to be determined

$ABCD$ is a parallelogram so

$\displaystyle \frac { 1 }{ 2 } \left( \overrightarrow { b } +\overrightarrow { d } \right) =\frac { 1 }{ 2 } \left( \overrightarrow { a } +\overrightarrow { c } \right)$

$\overrightarrow { d } =\overrightarrow { a } +\overrightarrow { c } -\overrightarrow { b } =8i+10j+12k-\left( i+3j+5k \right) \\ =7i+7j+7k=7\left( i+j+k \right)$

Let

$G$ be the centroid of a triangle

$ABC$ . If

$AB=a,AC=b$ then the bisector

$AG$ , in terms of vectors

$a$ and

$b$ is

0%
$\displaystyle \frac { 2 }{ 3 } \left( a+b \right)$
0%
$\displaystyle \frac { 1 }{ 6 } \left( a+b \right)$
0%
$\displaystyle \frac { 1 }{ 3 } \left( a+b \right)$
0%
$\displaystyle \frac { 1 }{ 2 } \left( a+b \right)$

Explanation

Take

$A$ as origin.

Then position vectors of

$A,B,C$ are given

$0,a$ and

$b$ respectively.

Then the centroid

$G$ has position vector

$\displaystyle \frac { 0+a+b }{ 3 } =\frac { a+b }{ 3 }$

$\displaystyle \therefore AG=\frac { a+b }{ 3 }$

What is the maximum number of components into which a vector can be
split ?

0%
2
0%
3
0%
4
0%
Infinite

The sum of the three vectors determined by the medians of a triangle directed from the vertices is

0%
$0$
0%
$1$
0%
$-1$
0%
$\displaystyle \frac{1}{3}$

Explanation

$\because D$ is middle point of

$BC$

$\displaystyle \therefore AD=\frac { AB+AC }{ 2 }$ ...(1)

$E$ is middle point of

$CA$

$\displaystyle \therefore BE=\frac { BA+BC }{ 2 }$ ....(2)

$F$ is middle point of

$AB$

$\displaystyle \therefore CF=\frac { CB+CA }{ 2 }$ ...(3)

Adding (1),(2) and (3), we get

$AD+BE+CF=0$

$P$ is any point on the circumcircle of

$\triangle ABC$ other than the vertices.

$H$ is the orthocenter of

$\triangle ABC,M$ is the mid-point of

$PH$ and

$D$ is the mid-point of

$BC$ . Then

0%
$AP$ is opposite side of $DM$
0%
$DM$ is parallel to $AP$
0%
$DM$ is perpendicular to $AP$
0%
None of these

Explanation

Let

$S$ be the circumcentre of

$\triangle ABC.$

Let

$\displaystyle \overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c }$ and

$\overrightarrow { p }$ be the orthocentre of vectors of

$A,B,C$ and

$P$ respectively with referentre to origin

$S.$

$\displaystyle \overrightarrow { SA } =\overrightarrow { a } ,\overrightarrow { SB } =\overrightarrow { b } ,\overrightarrow { SC } =\overrightarrow { c } ,\overrightarrow { SP } =\overrightarrow { p }$ and

$\displaystyle \left| \overrightarrow { p } \right| =\left| \overrightarrow { a } \right| =\left| \overrightarrow { b } \right| =\left| \overrightarrow { c } \right| =R=\\$ circumradius.

Now,

$\displaystyle \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\overrightarrow { SA } +\overrightarrow { SB } +\overrightarrow { SC }$

$\displaystyle =\overrightarrow { SA } +\overrightarrow { 2SD } =\overrightarrow { SA } +\overrightarrow { AH } =\overrightarrow { SH }$

$(\because H.G:GS=2:1$ where

$G$ is centroid

$)$

Position vector of

$\displaystyle H=\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }$

So, position vector of

$\displaystyle M=\frac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } +\overrightarrow { p } }{ 2 }$

$\overrightarrow { DM } =p.v$ of

$M-p.v.$ of

$\displaystyle D=\frac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } +\overrightarrow { p } }{ 2 } -\frac { \overrightarrow { b } +\overrightarrow { c } }{ 2 } =\frac { \overrightarrow { a } +\overrightarrow { p } }{ 2 }$

Now

$\displaystyle \overrightarrow { DM } .\overrightarrow { PA } =\left( \frac { \overrightarrow { a } +\overrightarrow { p } }{ 2 } \right) .\left( \frac { \overrightarrow { a } -\overrightarrow { p } }{ 2 } \right) =\frac { { \left| \overrightarrow { a } \right| }^{ 2 }-{ \left| \overrightarrow { p } \right| }^{ 2 } }{ 2 } =0\Rightarrow \overrightarrow { DM } \bot \overrightarrow { AP }$

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

$S$ is the circumcentre,

$O$ is the orthocenter of

$\triangle ABC,$ then

$SA+SB+SC=$

0%
$SO$
0%
$2SO$
0%
$OS$
0%
$2OS$

Explanation

From geometry

$2SD=AO,$ where

$D$ is the mid.

Point of

$BC.$

$\displaystyle \therefore SA+SB+SC=SA+SD+DB+SD+DC$

$=SA+2SD$

$=SA+AO=SO$

$\displaystyle \therefore SA+SB+SC=SO.$

$O$ and

$O'$ are circumcenter and orthocenter of a triangle

$ABC$ then

$\left( OA+OB+OC \right)$ equals

0%
$2OO'$
0%
$OO'$
0%
$O'O$
0%
$2O'O$

Explanation

according to the properties of triangle

circumcentre O in triangle ABC

the sum of distance from circumcentre to vertices of ABC is the distance from circumcentre to orthocentre

$(OA+OB+OC)=OO'$

$\displaystyle \vec{a},\vec{b},\vec{c},\vec{d},\vec{e},\vec{f}$ are position vectors of 6 points A, B, C, D, E & F respectively such that

$\displaystyle 3\vec{a}+4\vec{b}=6\vec{c}+\vec{d}=4\vec{e}+3\vec{f}=\vec{x}$ then

0%
$\displaystyle \overline{AB}$ is parallel to $\displaystyle \overline{CD}$
0%
line AB, CD and EF are concurrent
0%
$\displaystyle \frac{\vec{x}}{7}$ is position vector of the point dividing CD in ratio 1 : 6
0%
A, B, C, D, E & F are coplanar

Explanation

Given condition

$3\vec{a}+4\vec{b}=6\vec{c}+\vec{d}=4\vec{e}+3\vec{f}=\vec{x}$

It is clearly seen that sum of coefficient

$7=7=7=\vec{x}$

So lines are concurrent

position vector of c and d

$position\ vector=\dfrac{6c+d}{7}$

from above condition

$position\ vector=\dfrac{x}{7}$

Any vector in an arbitrary direction can always be replaced by two (or three)

0%
parallel vectors which have the original vector as their resultant.
0%
mutually perpendicular vectors which have the original vector as their resultant.
0%
arbitrary vectors which have the original vector as their resultant.
0%
it is not possible to resolve a vector.

$12$ coplanar non collinear forces (all of equal magnitude) maintain a body in equilibrium, then the angle between any two adjacent forces is:

0%
$\;15^{\circ}$
0%
$\;30^{\circ}$
0%
$\;45^{\circ}$
0%
$\;60^{\circ}$

Explanation

According to polygon law of vector addition, the 11 forces will be the sides of polygon in same order in direction and 1 forces will represent the resultant which is a side in opposite order of the polygon. Total angle enclosed by polygon is

$360^o$ . So, the required angle

$=\dfrac{360^o}{12}=30^o$

The value of

$\vec i \times (\vec a \times \vec i) + \vec j \times (\vec a \times \vec j) + \vec k \times (\vec a \times \vec k)$ is (where

$\vec i, \vec j, \vec k$ are unit vectors)

0%
$\vec a$
0%
$2 \vec a$
0%
0
0%
$- \vec a$

Explanation

$i \times (\bar a \times i) = (i . i) \bar a - (i. \bar a) i = \bar a - (i . \bar a)i$
so

$i \times (\bar a \times i) + j \times (\bar a \times j) + k \times (\bar a \times k)$

$= \bar a - (i . \bar a) i + \bar - (j . \bar a) j + \bar a - (k . \bar a) k$

$= 3\bar a - [(i . \bar a) i + (j . \bar a) j + (k . \bar a) k]$

$= 3 \bar a - \bar a = 2 \bar a$ .

If a and b are two non-zero and non-collinear vectors, then a + b and a - b are

0%
Linearly dependent vectors
0%
Linearly dependent and independent vectors
0%
Linearly independent vectors
0%
None of these

Let

$\vec{a}=\hat{i}+\hat{j}+3\hat{k}\;\&\;\vec{b}=2\hat{i}-3\hat{j}+4\hat{k}$ . If projection of

$\vec{a}$ on

$\vec{b}$ is

$\displaystyle\frac{k}{\sqrt{29}}$ , then the value of

$(k-2)$ is

0%
$9$
0%
$-9$
0%
$8$
0%
$6$

Explanation

GIVEN

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

$\vec{b}=2\hat{i}-3\hat{j}+4\hat{k}$

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

$Proj_{a}b=\dfrac{(\hat{i}+\hat{j}+3\hat{k})\cdot(2\hat{i}-3\hat{j}+4\hat{k})}{\left | \sqrt{2^2+(-3)^2+4^2} \right |}$

$Proj_{a}b=\dfrac{2-3+12}{\sqrt{4+9+16}}$

$Proj_{a}b=\dfrac{11}{ \sqrt{29}}$

comparing RHS with

$\dfrac{k}{ \sqrt{29}}$

$k=11$

$k-2=11-2$

$k-2=9$

If the position vectors of the vertices of a triangle be 6i + 4j + 5k, 4i + 5j + 6k and 5i + 6j + 4k then the triangle is

0%
Right angled triangle
0%
Equilateral triangle
0%
Isosceles triangle
0%
None of these

Explanation

We have,

Let the vertices of triangle is

$\overrightarrow{OA}=\left( 6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k} \right)$

$\overrightarrow{OB}=\left( 4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k} \right)$

$\overrightarrow{OC}=\left( 5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k} \right)$

Now,

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{AB}=\left( 4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k} \right)-\left( 6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k} \right)$

$\overrightarrow{AB}=4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k}-6\overrightarrow{i}-4\overrightarrow{j}-5\overrightarrow{k}$

$\overrightarrow{AB}=-2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$

$\overrightarrow{BC}=\left( 5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k} \right)-\left( 4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k} \right)$

$\overrightarrow{BC}=5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k}-4\overrightarrow{i}-5\overrightarrow{j}-6\overrightarrow{k}$

$\overrightarrow{BC}=\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$

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

$\overrightarrow{CA}=\left( 6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k} \right)-\left( 5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k} \right)$

$\overrightarrow{CA}=6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k}-5\overrightarrow{i}-6\overrightarrow{j}-4\overrightarrow{k}$

$\overrightarrow{CA}=\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k}$

Now,

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

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

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

Then,

$AB=BC=CA=\sqrt{6}$

Hence, It triangle is equilateral triangle.

In triangle

$ABC$ , which of the following is not true?

0%
$\vec {AB}+\vec {BC}+\vec {CA}=\vec {0}$
0%
$\vec {AB}+\vec {BC}-\vec {AC}=\vec {0}$
0%
$\vec {AB}+\vec {BC}-\vec {CA}=\vec {0}$
0%
$\vec {AB}-\vec {CB}+\vec {CA}=\vec {0}$

Explanation

${\textbf{Step 1: Using triangle law of vector addition, find the resultant of two vectors}}$

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

${\textbf{Step 2: Solve the vector sum and Compare with the given options}}$

$\Rightarrow \overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {AC} = \overrightarrow 0$

${\text{Or }}\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = \overrightarrow 0$

${\text{So, }}$ $\overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {CA} \ne \overrightarrow 0$

${\textbf{Hence, Option (C) is incorrect}}$

$\vec{a}=(\lambda\,x)\hat{i}+(y)\hat{j}+(4z)\hat{k},\,\vec{b}=y\hat{i}+x\hat{j}+3y\hat{k},\,\vec{c}=-z\hat{i}-2z\hat{j}-\begin{pmatrix}(\lambda+1)x\end{pmatrix}\hat{k}$ are sides of triangle as shown in figure, then value of

$\lambda$ is (where

$x,\,y,\,z$ are not all zero)

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

Explanation

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

$((\lambda\,x)\hat{i}+y\hat{j}+4z\hat{k})+(y\hat{i}+x\hat{j}+3y\hat{k})$

$=-z\hat{i}-2z\hat{j}-(\lambda+1)x\hat{k}$

$\Rightarrow\;\lambda\,x+y+z=0;\;\;x+y+2z=0$

$\&\;(\lambda+1)x+3y+4z=0$
For non-trivial solution,

$\begin{vmatrix}\lambda&1&1\\1&1&2\\(\lambda+1) & 3 & 4\end{vmatrix}=0$

$\Rightarrow \lambda(4-6)-(4-2(\lambda+1))+(3-(\lambda+1))=0$

$\Rightarrow -2\lambda-4+2\lambda+2+3-\lambda-1=0$

$\Rightarrow -\lambda=0\;\;\;$

$\Rightarrow\;\;\;\lambda=0$

The position vectors of

$P$ and

$Q$ are respectively

$a$ and

$b$ . If

$R$ is a point on

$PQ$ ,

$PQ$ such that

$PR=5PQ$ , then the position vector of

$R$ is

0%
$5b-4a$
0%
$5b+4a$
0%
$4b-5a$
0%
$4b+5a$

Explanation

Given condition

$PR=5PQ$

$R-P=5(Q-P)$

$R=5Q-5P+P$

$R=5Q-4P$

$R=5b-4a$

$|\vec {a}| = |\vec {b}| = 1$ and

$|\vec {a} + \vec {b}| = \sqrt {3}$ , then the value of

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

0%
$-21$
0%
$-\dfrac {21}{2}$
0%
$21$
0%
$\dfrac {21}{2}$

Explanation

$\because |\vec {a} + \vec {b}| = \sqrt {3}$

$\Rightarrow |\vec {a}|^{2} + |\vec {b}|^{2} + 2\vec {a} \cdot \vec {b} = 3$

$\Rightarrow \vec {a}\cdot \vec {b} = \dfrac {1}{2}$ ..... (i)

$(\because |\vec {a}| = |\vec {b}| = 1 given)$

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

$= 6|\vec {a}|^{2} + 15\vec {a}\cdot \vec {b} - 8\vec {a}\cdot \vec {b} - 20 |\vec {b}|^{2}$

$= 6 + 7\vec {a}\cdot \vec {b} - 20$

$= 6 + \dfrac {7}{2} - 20$ [from Eq. (i)]

$= \dfrac {7 - 28}{2} = -\dfrac {21}{2}$

$\vec { a } \cdot \hat { i } =4$ , then

$\left( \vec { a } \times \hat { j } \right) \cdot \left( 2\hat { j } -3\hat { k } \right)$ is equal to

0%
$12$
0%
$2$
0%
$0$
0%
$-12$

Explanation

$\left( \vec { a } \times \hat { j } \right) \cdot \left( 2\hat { j } -3\hat { k } \right) =\vec { a } \cdot \left\{ \hat { j } \times \left( 2\hat { j } -3\hat { k } \right) \right\}$

$=\vec { a } \cdot \left\{ -3\left( \hat { j } \times \hat { k } \right) \right\}$

$=-3\left( \vec { a } \cdot \hat { i } \right)$

$=-12$ (

$\because \vec { a } \cdot \hat { i } =4$ given)

Let

$ABC$ be a triangle whose circumcentre is at P. If the position vectors of

$A, B, C$ and P are

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

$\dfrac {\vec {a} + \vec {b} + \vec {c}}{4}$ respectively, then the position vector of the orthocentre of this triangle, is:

0%
$-\left (\dfrac {\vec {a} + \vec {b} + \vec {c}}{2}\right )$
0%
$\vec {a} + \vec {b} + \vec {c}$
0%
$\dfrac {(\vec {a} + \vec {b} + \vec {c})}{2}$
0%
$\vec {0}$

Explanation

O is orthocentre and G is centroid and C is circumcentre. G divides OC in

$2:1$ ratio.
Position vector of centriod

$\vec {G} = \dfrac {\vec {a} + \vec {b} + \vec {c}}{3}$
Position vector of circum center

$\vec {C} = \dfrac {\vec {a} + \vec {b} + \vec {c}}{4}$
Apply Section Formula,

$\vec {G} = \dfrac {2\vec {C} + \vec {R}}{3}$

$3\vec {G} = 2\vec {C} + \vec {R}$

$\vec {R} = 3\vec {G} - 2\vec {C} = (\vec {a} + \vec {b} + \vec {c}) - 2 \left (\dfrac {\vec {a} + \vec {b} + \vec {c}}{4}\right )$

$= \dfrac {\vec {a} + \vec {b} + \vec {c}}{2}$

Let

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

$\overrightarrow {c}$ be vectors with magnitudes 3, 4 and 5 respectively and

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

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

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

Explanation

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

dot product with

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

$\vec{c}$ both sides

$\vec{a}.(\vec{a}+\vec{b}+\vec{c}) = \vec{0}\Rightarrow (\vec{a})^{2}+\vec{a}.\vec{b}+\vec{a}.\vec{c} = \vec{0}.$

Similarly,

$\vec{b},\vec{a}+(\vec{b})^{2}+(\vec{b}.\vec{c}) = \vec{0}; \vec{c}.\vec{a}+\vec{b}.\vec{c}+(\vec{c})^{2} = \vec{0}$

adding all 3 equation,

$(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}) = \vec{0}$

$2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}) = -[3^{2}+4^{2}+5^{2}] = -50$

$\boxed { \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a} = -25}$

1126187_460595_ans_5f07d573e8ef44358c670bed8c470511.JPG

Let

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

$\vec {c} = \vec {i} + \vec {j} - \vec {k}$ . A vector in the plane of

$\vec {a}$ and

$\vec {b}$ has projection

$\dfrac {1}{\sqrt {3}} \ on\ \vec {c}$ . Then, one such vector is

0%
$4\vec {i} + \vec {j} - 4\vec {k}$
0%
$3\vec {i} + \vec {j} - 3\vec {k}$
0%
$4\vec {i} - \vec {j} + 4\vec {k}$
0%
$2\vec {i} + \vec {j} - 2\vec {k}$

Explanation

Let

$\vec { v }$ be the vector then

$\vec { v } =\vec { a } +\lambda \vec { b }$

$\therefore \vec { v } =\left( 1+\lambda \right) \hat { i } +\left( 2-\lambda \right) \hat { j } +\left( 1+\lambda \right) \hat { k }$

Now

$\left| \dfrac { \vec { v } \cdot \vec {c } }{ \left| \vec { c } \right| } \right| =\dfrac { 1 }{ \sqrt { 3 } }$

$\Rightarrow \dfrac { \left| 1+\lambda +2-\lambda -1-\lambda \right| }{ \sqrt { 3 } } =\dfrac { 1 }{ \sqrt { 3 } }$

$\Rightarrow 2-\lambda =\pm 1$

$\Rightarrow \lambda =3$ or

$1$

$\therefore \vec { v } =4\hat { i } -\hat { j } +4\hat { k }$ or

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

Find the correct vectorial relationship with the help of the figure above.

0%
$\vec {x} + \vec {y} = \vec {z}$
0%
$\vec {y} + \vec {z} = \vec {x}$
0%
$\vec {x} + \vec {z} = \vec {y}$
0%
$\vec {z} - \vec {x} = \vec {y}$
0%
$\vec {z} - \vec {y} = \vec {x}$

Explanation

If we observe clearly, we can see that

$y$ is resultant of

$x$ and

$z$ .
So, we get

$y = x+z$ .

How much does a watch lose per day, if its hands coincide every

$64$ minutes?

0%
$32\cfrac { 8 }{ 11 } min.$
0%
$31\cfrac { 8 }{ 11 } min.$
0%
$32\cfrac { 3 }{ 11 } min.$
0%
$None\ of\ these$

Explanation

$55min.$ spaces are covered in

$60min.$

$60min.$ spaces are covered in

$\left( \cfrac { 60 }{ 55 } \times 60 \right) min=65\cfrac { 5 }{ 11 } min.$

Loss in

$64min.=\left( 65\cfrac { 5 }{ 11 } -64 \right) =\cfrac { 16 }{ 11 } min.$

Loss in

$24hrs.=\left( \cfrac { 16 }{ 11 } \times \cfrac { 1 }{ 64 } \times 24\times 60 \right) min.=32\cfrac { 8 }{ 11 } min$

$ABCDEF$ is a regular hexagon whose centre is

$O$ . The

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

0%
$2\overline { AO }$
0%
$3\overline { AO }$
0%
$5\overline { AO }$
0%
$6\overline { AO }$

Explanation

Since it is a regular hexagon,

$\overline { AB } =\overline { ED } ,\overline { AF } =\overline { CD }$

Now,

$\overline { AB } +\overline { AC } +\overline { AD } +\overline { AE } +\overline { AF } =\overline { AE } +\overline { ED } +\overline { AC } +\overline { CD } +\overline { AD } =3\overline { AD }$

Now, since

$O$ is the centre of the hexagon,

$O$ will be the midpoint of

$AD$

$\therefore 3\overline{AD}=6\overline { AO }$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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