MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 3

find the coordinate of the tip of the position vector which is equivalent to

$\vec{AB}$ , where the coordinates of A and B are (-1, 3) and (-2, 1) respectively.

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(+1,+2)
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(+1,-2)
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(-1,+2)
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(-1,-2)

Explanation

$A(-1, 3)$

$B(-2, 1)$

'O' be the origin

$\vec{OA}=-\hat{i}+3\hat{j}$ ;

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

$\vec{AB}=\vec{OB}-\vec{OA}=(-2\hat{i}+\hat{j})-(-\hat{i}+3\hat{j})$

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

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

$\therefore$ Coordinates

$=(-1, -2)$ .

If the position vectors of the points

$A(3,4),B(5, -6)$ and

$C(4,-1)$ are

$\vec{a}, \vec{b}, \vec{c}$ respectively, compute

$\vec{a}+2\vec{b}-3\vec{c}.$

0%
$-5\hat{i}-1\hat{j}$
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$\hat{i}-5\hat{j}$
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$\hat{i}+5\hat{j}$
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none of these

Explanation

Position vectors of points

$A(3, 4), B(5, -6), C(-4, -1)$ are

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

$\vec{a}=3\hat{i}+4\hat{j}$

$\vec{b}=5\hat{i}-6\hat{j}$

$\vec{c}=4\hat{i}-\hat{j}$

$\vec{a}+2\vec{b}-3\vec{c}=3\hat{i}+4\hat{j}+2(5\hat{i}-6\hat{j})-3(4\hat{i}-\hat{j})$

$=3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}-12\hat{i}+3\hat{j}$

$=\hat{i}-5\hat{j}$

$\vec{a}+2\vec{b}-3\vec{c}=\hat{i}-5\hat{j}$ .

$4 \hat{i} + 7 \hat{j} + 8 \hat{k}, 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and

$2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ are the position vectors of the vertices

$A, B$ and

$C$ respectively, of the triangle

$ABC$ , the position vector of the point where the bisector of angle

$A$ meets

$BC$ , is

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$\frac{2}{3} (-6 \hat{i} - 8 \hat{j} - 6 \hat{k})$
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$\frac{2}{3} (6 \hat{i} + 8 \hat{j} + 6 \hat{k})$
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$\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$
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$\frac{1}{3} ( 5 \hat{j} + 12 \hat{k})$

Explanation

Suppose the bisector of angle

$A$ meets

$BC$ at

$D$ . Then

$AD$ divides

$BC$ in the ratio

$AB : AC$ .

So, P.V. of

$D = \frac{\mid \overrightarrow{AB}\mid (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + \mid \overrightarrow{AC}\mid (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{\mid \overrightarrow{AB}\mid + \mid \overrightarrow{AC}\mid}$

But

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

$\overrightarrow{AC} = - 2 \hat i - 2 \hat j - \hat k$

$\overrightarrow{AB} = 6$ and

$\overrightarrow{AC} = 3$

$\therefore P.V. of D = \frac{6(2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + 3(2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{6 + 3}$

$= \frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$

If vectors

$\overrightarrow{AB} = -3 \hat{j} + 4 \hat{k}$ and

$\overrightarrow{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a

$\Delta ABC$ , then the length of the median through

$A$ is

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$\sqrt{14}$
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$\sqrt{18}$
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$\sqrt{29}$
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$5$

Explanation

$\overrightarrow{AB} + \overrightarrow{AC} = 2 \overrightarrow{AD}$

$\therefore \overrightarrow{AD} = \frac{1}{2} \{ (-3 \hat{j} + 4 \hat{k}) + (5 \hat{i} - 2 \hat{j} + 4 \hat{k}) \}$

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

Length of

$AD = \sqrt{1 + 1 + 16} = \sqrt 18$

1655584_1782257_ans_11788933976a426ab55b729a17394571.png

Let

$ABC$ be a triangle, the position vectors of whose vertices are respectively

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

$2 \hat{i} + 4 \hat{j} + 3 \hat{k}$ . Then

$\Delta ABC$ is

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isosceles
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equilateral
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right angled
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none of these

Explanation

$\vec{BC} = \vec{OC} - \vec{OB} = 4 \hat{i} + 2 \hat{j} - 4\hat{k}$

$\vec{AB} = 3 \hat{i} - 3 \hat{k}, \vec{AC} = \hat{i} + 2\hat{j} - 7 \hat{k}$

$BC^2 = 36, AB^2 = 18, AC^2 = 54$

Clearly,

$AC^2 = BC^2 + AB^2$

$\therefore \angle B = 90^o$

$\vec a$ is parallel to

$\vec b \times \vec c$ , then

$(\vec a \times \vec b) \cdot (\vec a \times \vec c)$ is equal to

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$\mid \vec a \mid^2 (\vec b \cdot \vec c)$
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$\mid \vec b \mid^2 (\vec a \cdot \vec c)$
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$\mid \vec c \mid^2 (\vec a \cdot \vec b)$
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none of these

Explanation

$(\vec a \times \vec b) \cdot (\vec a \times \vec c) = (\vec a \times \vec b) \cdot \vec u = \vec a \times \vec c$

$\Rightarrow \vec a \cdot (\vec b \times \vec u) = \vec a \cdot [\vec b \times (\vec a \times \vec c)]$

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

$= \vec a \cdot (\vec b \cdot \vec c) \vec a$

$(\because \vec a \cdot \vec b = 0)$

$= \mid \vec a\mid^2 (\vec b \cdot \vec c)$

$\mid \vec a \mid = 2$ and

$\mid \vec b \mid = 3$ and

$\vec a \cdot \vec d = 0$ , then

$(\vec a \times (\vec a \times (\vec a \times (\vec a \times \vec b ))))$ is equal to

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$48 \hat b$
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$- 48 \hat b$
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$48 \hat a$
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$48 \hat a$

Explanation

$(\vec a \times (\vec a \times (\vec a \times (\vec a \times \vec b )))) = (\vec a \times ( \vec a \times ((\vec a \cdot \vec b) \vec a - (\vec a \cdot \vec b) \vec b)))$

$= (\vec a \times (\vec a \times (- 4 \vec b)))$

$= -4 (\vec a \times(\vec a \times \vec b))$

$= -4((\vec a \cdot \vec b) \vec a - (\vec a \cdot \vec a)\vec b)$

$= -4(-4 \vec b) = 16 \vec b = 48 \hat b$

Figure shows

$ABCDEF$ as a regular hexagon. What is the value of

$\vec{AB}+\vec{AC}+\vec{AD}+\vec{AE}+\vec{AF}$

0%
$\vec{AO}$
0%
$2\vec{AO}$
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$4\vec{AO}$
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$6\vec{AO}$

Explanation

the value of

$\vec{AB}+\vec{AC}+\vec{AD}+\vec{AE}+\vec{AF}$

$6\vec{AO}$ AS

$\vec {AC}=\vec{AO}+\vec{OC}$

and

$\vec{AD}=2\vec{AO}$

similarly on adding all we get

$6\vec{AO}$

1709526_1811900_ans_a23664da9ec349acb8c03eeb6459cfba.png

Let

$\mathrm{A}\mathrm{B}\mathrm{C}$ be a triangle and let

$\mathrm{D},\mathrm{E}$ be the midpoints of the sides

$\mathrm{A}\mathrm{B},\mathrm{A}\mathrm{C}$ respectively,then

$\hat { BE } +\hat { DC } =$

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$\hat { BC }$
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${ \dfrac { 1 }{ 2 } }\hat { BC }$
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${ \dfrac { 3 }{ 2 } \hat { BC } }$
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$\dfrac { 3 }{ 4 } \hat { BC }$

Explanation

Since,

$D$ and

$E$ are the mid points of the sides

$AB$ and

$AC$

Therefore,

$OD=\dfrac{OA+OB}{2}$ and

$OE=\dfrac{OA+OC}{2}$

$Now, BE+DC\\=OE-OB+OC-OD=OC-OB+\dfrac{OC-OB}{2}\\=\dfrac{3(OC-OB)}{2}=\dfrac{3BC}{2}$

Let

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

$\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector

$\vec{v}$ in the plane of

$\vec{a}$ and

$\vec{b}$ , whose projection on

$\vec{c}$ is

$\displaystyle \dfrac{1}{\sqrt{3}}$ , is given by

$;$

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

Explanation

Let

$\overrightarrow { v } =\lambda \overrightarrow { a } +\mu \overrightarrow { b }$

$\Rightarrow\overrightarrow { v } =(\lambda +\mu )\hat { i } +(\lambda -\mu )\hat { j } +(\lambda +\mu )\hat { k }$

Projection of

$\overrightarrow { v }$ on

$\displaystyle\overrightarrow { c }=\dfrac { \overrightarrow { v } .\overrightarrow { c } }{ \left| \overrightarrow { c } \right| } = \dfrac { 1 }{ \sqrt { 3 } }$

$\Rightarrow\displaystyle\dfrac { (\lambda +\mu )-(\lambda -\mu )-(\lambda +\mu ) }{ \sqrt { 3 } } =\dfrac { 1 }{ \sqrt { 3 } }$

$\Rightarrow\mu -\lambda =1$
or

$\mu =\lambda +1$

$\Rightarrow\overrightarrow { v } =(2\lambda +1)\hat { i } -\hat { j } +(2\lambda +1)\hat { k }$
For

$\lambda =1, \overrightarrow { v }=3\hat { i } -\hat { j }+3\hat { k }$

The position vectors of

$A,B,C$ are

$\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+5\overline{j}+\overline{k}$ . Then the position vector of the circumcentre of the triangle

$ABC$ is

0%
$3\hat{i}+2\hat{j}+\hat{k}$
0%
$\displaystyle \frac{1}{2}(6\hat{i}+\hat{j}+\hat{k})$
0%
$\displaystyle \frac{1}{2}(5\hat{i}+6\hat{j}+2\hat{k})$
0%
${\dfrac{1}{2}(9\overline{i}}+7\overline{j}+3\overline{k})$

Explanation

Position vector of $A$ is $\hat i+\hat j+\hat k$ , gives coordinates as $(1,1,1)$ .

Position vector of

$B$ is

$4\hat i+\hat j+\hat k$ , gives coordinates as

$(4,1,1)$ .

Position vector of

$C$ is

$4\hat i+5\hat j+\hat k$ , gives coordinates as

$(4,5,1)$ .

$\vec{AB} = \vec B - \vec A = 3\hat i$

$\vec {BC} = \vec C - \vec B = 4\hat j$

$\vec {AC} = \vec C - \vec A = 3\hat i+4\hat j$

Here, we can see that

$\triangle ABC$ is a right angled triangle with right angle at

$B$ .

Hence, circumcenter will be the midpoint of

$AC$ .

Mid point of

$AC = D\equiv\left(\dfrac{1+4}2 , \dfrac{1+5}2, \dfrac{1+1}2\right) = \left(\dfrac52,3,1\right)$

So, position vector of

$D$ will be

$\dfrac12(5\hat i+6\hat j+2\hat k)$

which is a circumcenter.

The vector sum of

$N$ coplanar forces, having magnitude of

$F$ , when each force is making an angle of

$\displaystyle \frac{2\pi}{N}$ with that preceding it, is:

0%
$F$
0%
$\displaystyle \frac{NF}{2}$
0%
$NF$
0%
$0$

Explanation

If there are

$N$ forces of magnitude

$F$ and each force makes an angle of

$\dfrac { 2 \pi}{N}$ with the preceding one, then these forces form the sides of a

$N$ sided regular polygon thus making the resultant of all the forces equal to zero.

If the vectors

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

$\overline{b}=-\overline{i}+3\overline{j}+4\overline{k},\ \overline{c}=4\overline{i}-2\overline{j}-6\overline{k}$ form the sides of the triangle then length of the median bisecting the vector

${c}$ is

0%
$\sqrt{12}$ units
0%
$\sqrt{6}$ units
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$2\sqrt{6}$ units
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$2\sqrt{3}$ units

Explanation

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

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

$\vec c = 4\hat i-2\hat j-6\hat k = \vec{AB}$

$|\vec a|^2 = (3^2+1^2+2^2) = 14$

$|\vec b|^2 = 1^2+3^2+4^2 = 26$

$|\vec c|^2 = 4^2+2^2+6^2 = 56$

Length of median through vertex

$C$ passing through

$AC$ :

$L = \dfrac12\sqrt{2(a^2+b^2)-c^2}$

$L = \dfrac12\sqrt{2(14+26) - 56} = \dfrac12\sqrt{24} = \sqrt6$

$\mathrm{O}$ is the circumcentre and

$\mathrm{O}^{'}$ is the orthocentre of a triangle

$\mathrm{A}\mathrm{B}\mathrm{C}$ and if

$\mathrm{A}\mathrm{P}$ is the circumdiameter then

$\vec{\mathrm{A}\mathrm{O}}+\vec{\mathrm{O}^{'}\mathrm{B}}+\vec{\mathrm{O}^{'}\mathrm{C}}=$

0%
$\vec{\mathrm{O}\mathrm{A}}$
0%
$\vec{\mathrm{O}^{'}\mathrm{A}}$
0%
$\vec{\mathrm{A}\mathrm{P}}$
0%
$\vec{\mathrm{A}\mathrm{O}}$

Explanation

$\left | \vec{AP} \right |=2\left | AO \right |$

$=O'B+O'C+O'A-O'A+AO'$

$=(\bar{A}+\bar{B}+\bar{C}-3 \ \bar{O'})+2\ \overline{AO'}$
So use

$3\ \overline{OS}=2\ \overline{OC}$
where

$S$ is centroid

$C$ is circumcentre

$O$ is orthocenter

$=3(\dfrac{\bar{A}+\bar{B}+\bar{C}-\bar{O'}}{3})+2\overline{AO'}$

$=2\ \overline{O'O}+2\ \overline{AO'}$

$=2\ \overline{AO}$

$=\overline{AP}$

Five equal forces each of 20N are acting at a point in the same plane. If the angles between them are same, then the resultant of these forces is:

0%
0
0%
40N
0%
20N
0%
$20\sqrt{2}$

If there are 11 vectors each having a magnitude equal to

$| \vec{p} |$ and if each side of polygon subtends an angle

$30^{0}$ at the centre of the polygon. Then the resultant is

0%
$\vec{p}$
0%
0
0%
$\frac{\vec{p}}{2}$
0%
2 $\vec{p}$

Explanation

$12$ vectors at $30^0$ can make resultant $\overrightarrow{0}$ due to circular symmetry.

Vector sum of $11$ vector $+ 12$ vector $=0$

$R+\overrightarrow{p}$ $=0$

$R=-$ $\overrightarrow{p}$

Hence, resultant vector will be opposite direction of $R$ .

$\mathrm{If}$

$\vec{AD},\ \vec{BE},\ \vec{CF}$ are medians of an equilateral triangle

$\mathrm{A}\mathrm{B}\mathrm{C}$ , then

$\vec{AD}+\vec{BE}+\vec{CF}$ equals to

0%
$\vec{AB}+\vec{BC}+\vec{CA}$
0%
a zero vector
0%
both $1$ and $2$
0%
$2\vec{AF}+3\vec{BF}$

Explanation

The question wants to ask what is the vector sum of these

$23$ median vectors. The answer is correct. Let the vectors of vertices be

$a, b$ and

$0$ then the median vectors are

$(a+b)/2, b/2 - a$ and

$a/2 - b$
The sum is a zero vector. and

$a+ b-a +(-b)$ which is vector sum of

$\vec{AB}, \vec{BC}$ and

$\vec{CA}$ is also

$0$ .

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

$\overrightarrow { c }$ are three non-zero vectors such that

$\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } ,$ then

0%
$\overrightarrow { b } =\overrightarrow { c }$
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$\overrightarrow { a } \bot \overrightarrow { b } ,\overrightarrow { c }$
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$\overrightarrow { a } \bot \overrightarrow { b } -\overrightarrow { c }$
0%
either $\overrightarrow { a } \bot (\overrightarrow { b } -\overrightarrow { c })$ or $\overrightarrow { b } =\overrightarrow { c }$

Explanation

Here If

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

$\vec{c}$
     are   three non-zero vectors   such that

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

$\vec{a} .\vec{c}$

$\vec{a} .\vec{b}$ -

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

$\vec{a}.(\vec{b} - \vec{c}) =0$

$\vec{a} =0 , (\vec{b} - \vec{c}) =0$

$\vec{a} =0$ or

$\vec{b} = \vec{c}$
Therefore from this , we get
Either

$\vec{a}\perp (\vec{b} - \vec{c})$ or

$\vec{b} = \vec{c}$
Hence Option

$(D)$

Let

$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ be a parallelogram and let

$\mathrm{L}$ and

$\mathrm{M}$ be the midpoints of the sides

${ BC }$ and

${ CD }$ respectively. Then

$\vec { AL } +\vec { AM } =$

0%
$\vec { AC }$
0%
${ \dfrac { 2 }{ 3 } }\vec { AC }$
0%
$\dfrac { 3 }{ 2 } \vec { AC }$
0%
$2\vec { AC }$

Explanation

$BL=LC$

$CM=DM$

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

$\vec{AD}=\vec{BC}=\vec{b}$

$\vec{AL}=\vec{AB}+\vec{BL}$

$\vec{AM}=\vec{AD}+\vec{DM}$

$\vec{AL}+\vec{AM}=\vec{AB}+\vec{\dfrac{BC}{2}}+\vec{AD}+\vec{\dfrac{DC}{2}}$

$=\dfrac{3}{2}\left ( \vec{AB}+\vec{BC} \right )$

$=\dfrac{3}{2}\vec{AC}$

Let

$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ be a trapezium and let

$\mathrm{P},\mathrm{Q}$ be the midpoints of the nonparallel sides

$\mathrm{A}\mathrm{D},\ \mathrm{B}\mathrm{C}$ respectively. Then

$\vec { PQ }$

0%
$\vec { AB } +\vec { DC }$
0%
${ \dfrac { 1 }{ 4 } }(\vec { AB } +\vec { BC } )$
0%
$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { DC } )$
0%
$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { BC } )$

Explanation

$AP=PD$ (

$P$ is mid point)

$BQ=QC$ (

$Q$ is mid point)

Let

$\vec{a},\vec{b},\vec{c}, \vec{d}$ are position vector of

$ABCD$ respectively
position vector of

$P = \dfrac{\vec{a}+\vec{d}}{2}$
position vector of

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

$\vec{PQ}=\dfrac{1}{2}(\vec{b}-\vec{a}-\vec{d}+\vec{c})$

$=\dfrac{1}{2}(\vec{AB}-\vec{CD})$

$=\dfrac{1}{2}(\vec{AB}+\vec{DC})$

${A}{B}{C}{D}$ is a quadrilateral,

$E$ is the point of intersection of the line joining the middle points of the opposite sides. lf

$O$ is any point, then

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

0%
$4\hat { OE }$
0%
$3\hat { OE }$
0%
$2\hat { OE }$
0%
$\hat { OE }$

Explanation

Let

$PV$ (position vector) of

$A,B,C,D$ be

$\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.

$PV$ of

$P$

$=$

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

$PV$ of

$Q$

$=$

$\dfrac{\vec{c}+\vec{d}}{2}$

$PV$ of

$v$

$=$

$\dfrac{\vec{a}+\vec{d}}{2}$

$PV$ of

$u$

$=$

$\dfrac{\vec{b}+\vec{b}}{2}$

$\vec{PQ}=\dfrac{1}{2}(\vec{d}-\vec{a}+\vec{c}-\vec{b})$

$\vec{uv}=\dfrac{1}{2}(\vec{a}-\vec{b}+\vec{d}-\vec{c})$
So,

$E=\dfrac{\vec{a}+\vec{b}+\vec+{c}+\vec{d}}{4}$
Let

$O$ be origin, so

$\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=\vec{a}+\vec{b}+\vec{c}+\vec{d}$
Therefore,

$4\vec{OE}=\vec{a}+\vec{b}+\vec{c}+\vec{d}$

$\overline{DA}=\overline{a};\overline{AB}=\overline{b}$ and

$\overline{CB}=k\overline{a}$ where

$k>0$ and

$x, y$ are the midpoints of

$DB$ and

$AC$ respectively such that

$|\overline{a}|=17$ and

$|\overline{XY}|=4$ , then

$\mathrm{k}=$

0%
$\displaystyle \dfrac{8}{17}$
0%
$\displaystyle \dfrac{9}{17}$
0%
$\displaystyle \dfrac{11}{17}$
0%
$\displaystyle \dfrac{4}{17}$

Explanation

$AY=CY$ .......

$[y$ is the midpoint of

$AC]$

$DX=BX$ .......

$[x$ is the midpoint of

$DB]$

Let D be the origin
Position vector(P.V.) of

$A=\vec{a}$

P.V. of

$B=\vec{a}+\vec{b}$ .......

$[\because \overline{AB}=\bar{b}]$
P.V. of

$C=-(k-1)\vec{a}+\vec{b}$ .......

$[\because \overline{CB}=k\bar{a}]$
P.V. of

$x=\dfrac{\vec{a}+\vec{b}}{2}$
P.V. of

$y=\dfrac{-k\vec{a}+2\vec{a}}{2}+\dfrac{\vec{b}}{2}$

$\vec{XY}=\dfrac{\vec{a}-k\vec{a}}{2}$

$\left | \vec{XY} \right |=\dfrac{\left | a \right |}{2}\sqrt{(1-k)^{2}}$

$\implies \dfrac{8}{17}=1-k$

$\implies k=\dfrac{9}{17}$

If the vectors

$4\hat{i}-7\hat{j}-2\hat{k},\: \hat{i}+5\hat{j}-3\hat{k},\: 3\hat{i}-\lambda\hat{j}+\hat{k}$ form a triangle then

$\lambda=$

0%
$6$
0%
$-6$
0%
$12$
0%
$-1$

Explanation

Triangle follow Addition Law

$\hat{i}+5\hat{j}-3\hat{k}+3\hat{i}-\lambda \hat{j}+\hat{k}=4\hat{i}-7\hat{j}-2\hat{k}$

$(5-\lambda )=-7$

$\lambda =12$

The incentre of the triangle formed by the points

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

$4{ \hat { i } }+{ \hat { j } }+{ \hat { k } }$ , and

$4{ \hat { i } }+5{ \hat { j } }+\hat { k }$ is

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

Explanation

Position vector of $A$ is $\hat i+\hat j+\hat k$ , gives coordinates as $(1,1,1)$ .

Position vector of

$B$ is

$4\hat i+\hat j+\hat k$ , gives coordinates as

$(4,1,1)$ .

Position vector of

$C$ is

$4\hat i+5\hat j+\hat k$ , gives coordinates as

$(4,5,1)$ .

$\vec{AB} = \vec B - \vec A = 3\hat i$

$\vec {BC} = \vec C - \vec B = 4\hat j$

$\vec {AC} = \vec C - \vec A = 3\hat i+4\hat j$

Here, we can see that

$\triangle ABC$ is a right angled triangle with right angle at

$B$ .

$|\vec{AB}| = \sqrt{3^2} = 3$

$|\vec{BC}| = \sqrt{4^2} = 4$

$|\vec{AC}| = \sqrt{3^2+4^2} = 5$

Point of Intersection of angle bisectors can be found using formula:

$I = \dfrac{|\vec{BC}|\times\vec A + |\vec{CA}|\times\vec B + |\vec{AB}|\times\vec C}{|\vec{AB}| + |\vec{BC}| + |\vec{CA}|}$

$I = \dfrac{4\times(\hat i+\hat j+\hat k) + 5\times(4\hat i+\hat j+\hat k) + 3\times(4\hat i+5\hat j+\hat k)}{3+4+5}$

$\Rightarrow I = \dfrac{1}{12}(36\hat i + 24\hat j+12\hat j) = 3\hat i+2\hat j+\hat k$

Hence, incenter will be

$3\hat i+2\hat j +\hat k$

$4{\vec{i}}+7\vec{j}+8\vec{k} , 2\vec{i}+3\vec{j}+4\vec{k}$ and

$2\vec{i}+5\vec{j}+7\vec{k}$ are the position vectors of the vertices

$\mathrm{A},\mathrm{B}$ and

$\mathrm{C}$ of

$\triangle \mathrm{A}\mathrm{B}\mathrm{C}$ , the position vector of

$D$ the point where the bisector of

$\angle A$ meets

$\mathrm{B}\mathrm{C}$ is

0%
${\dfrac{2}{3}}(-6\hat {i}-8\hat {j}-6\hat {k})$
0%
${\dfrac{2}{3}}(6\hat {i}+8\hat{j}+6\hat{k})$
0%
${\dfrac{1}{3}}(6\hat{i}+13\hat{j}+18\hat{k})$
0%
$2(\hat{i}+\hat{j}+\hat{k})$

$\hat { AB } =-3{ \hat { i } }+4{ \hat { k } }$ and

$\hat { BC } =-\overline { i } -2\overline { k }$ are the sides of the triangle

$ABC$ then the length of the median

$AM$ is

0%
$\sqrt{\dfrac{25}{2}}$
0%
$\sqrt{\dfrac{45}{2}}$
0%
$\displaystyle \dfrac{\sqrt{65}}{2}$
0%
$\displaystyle \dfrac{\sqrt{85}}{2}$

Explanation

Given

$\overrightarrow{AB}=-3\hat{i}+4\hat{k}$

$\overrightarrow{BC}=-\hat{i}-2\hat{k}$

Let

$\overrightarrow{A} = 0\hat{i}$

$+0\hat{j}$

$+0\hat{k}$ ,

$\overrightarrow{AB} =\overrightarrow{B}-\overrightarrow{A}$

$\Rightarrow \overrightarrow{B} =-3\hat{i} +4\hat{k}$

$\overrightarrow{BC} =\overrightarrow{C}-\overrightarrow{B}$

$\overrightarrow{C} = -3\hat{i}+4\hat{k}+(-\hat{i}-2\hat{k})$ ,

$\overrightarrow{C} = -4\hat{i}+2\hat{k}$

Median

$\overrightarrow{AM} = \dfrac{\overrightarrow{B}+\overrightarrow{C}}{2}$

$\overrightarrow{AM}= -\dfrac{7}{2}\hat{i} + 3\hat{k}$

$\left | \overrightarrow{AM} \right | =\sqrt{(\dfrac{7}{2})^2+3^2}$

$\left | \overrightarrow{AM} \right | =\sqrt{\dfrac{85}{4}}$

$\left | \overrightarrow{AM} \right | =\dfrac{\sqrt{85}}{2}$

Hence, option D is correct.

$A=(-3,2,5), B=(-3,4,5)$ and

$C=(-3,4,7)$ are the position vectors of vertices of

$\Delta ABC$ then its circumcentre is

0%
$(-3,3,5)$
0%
$(-3,3,6)$
0%
$(-3,4,6)$
0%
$(-3,4,7)$

Explanation

$\overrightarrow{AB}=2\hat{j}$

$\overrightarrow{AC}=2\hat{j}+2\hat{k}$

$\overrightarrow{BC}=2\hat{k}$
Triangle

$ABC$ is right angled at angle

$B$ .
Hence, the circumcentre is mid point of

$AC$ . Let the circumcentre be

$D$ .

$=\frac{1}{2}(-6,6,12)$

$=(-3,3,6)$

$\vec{a}$ and

$\vec{b}$ are two non-parallel unit vectors and the vector

$\alpha\vec{a}+\vec{b}$ bisects the internal angle between

$\vec{a}$ and

$\vec{b}$ , then

$\alpha$ is

0%
$1$
0%
$\dfrac{1}{2}$
0%
$2$
0%
$5$

Explanation

Given

$\vec{a}\ and\ \vec{b}\ are\ non-parallel\ unit\ vector$

$\left | \vec{a} \right |=1$ and

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

Internal angle bisector of

$\vec{a}$ and

$\vec{b}$ is

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

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

$\left | \vec{a} \right |=1$ and

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

and Given bisector is

$\alpha\vec{a}+\vec{b}$

Comparing Both the bisector

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

from above eq

$\alpha=1$ .

Two forces act at the vertex

$\mathrm{A}$ of quadrilateral

$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ represented by

$\overline{AB},\ \overline{AD}$ and two at

$\mathrm{C}$ represented by

$\overline{CD}$ and

$\overline{CB}$ . If

$\mathrm{E},\ \mathrm{F}$ are mid points of

$\overline{AC}$ and

$\overline{BD}$ respectively, then their resultant is

0%
$\overline{EF}$
0%
$2\overline{EF}$
0%
${\dfrac{3}{2}\overline{EF}}$
0%
$4\overline{EF}$

Explanation

Let position vector of

$A,B,C,D$ are

$\vec{a},\vec{b},\vec{c},\vec{d}$
Position Vector of

$E=\dfrac{\vec{a}+\vec{c}}{2}$
Position Vector of

$F=\dfrac{\vec{b}+\vec{d}}{2}$

So,

$\vec{AB}=\vec{b}-\vec{a}$ ,

$\vec{AD}=\vec{d}-\vec{a}$ ,

$\vec{CD}=\vec{d}-\vec{c}$ and

$\vec{CB}=\vec{b}-\vec{c}$
Resultant=

$\vec{AB}+\vec{AD}+\vec{CD}+\vec{CB}$

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

$=2(2F-2E)$

$=4\vec{FE}$

If point

$O$ is the centre of a circle circumscribed about a triangle

$ABC$ . Then

$\overline{OA}\sin 2A+\overline{OB}\sin2B+\overline{OC}\sin 2C=$

0%
$(\overline{OA}+\overline{OB}+\overline{OC})\sin 2A$
0%
$(\overline{OA}+\overline{OB}+\overline{OC})\cos 2A$
0%
$\overline{0}$
0%
$(\overline{OA}+\overline{OB}+\overline{OC})\tan 2A$

Explanation

In order to make the problem easier and take less time to solve it, we take a simpler case and take the triangle to be an equilateral one.
Thus

$OA = OB = OC$ and

$sin 2A = sin 2B = sin 2C = sin (120^0) = \dfrac{\sqrt {3}}{2}$
So after solving we get

$\overline {OA}$ +

$\overline{OB}$ +

$\overline{OC}$

$= 0.$
Hence the answer comes out to be

$0.$

Let

$G$ and

$G^{1}$ be the centroids of the triangles

$ABC$ and

$A^{1}B^{1}C^{1}$ respectively, then

$AA^{1}+BB^{1}+CC^{1}$ is equal to

0%
$2GG^{1}$
0%
$3G^{1}G$
0%
$3GG^{1}$
0%
$\displaystyle \dfrac{3}{2}GG^{1}$

Explanation

$G=\dfrac{A+B+C}{3}\Rightarrow A+B+C=3G$ ......(1)

and

$G^1=\dfrac{A^1+B^1+C^1}{3}\Rightarrow A^1+B^1+C^1=3G^1$ ......(2)

$\therefore AA^1+BB^1+CC^1 = (A-A^1)+(B-B^1)+(C-C^1)$

$=(A+B+C)-(A^1+B^1+C^1)$

$= 3G-3G^1=3(G-G^1)=3GG^1$ , Using (1) and (2)

The ratio in which

$\overline{i}+2\overline{j}+3\overline{k}$ divides the join

$\mathrm{o}\mathrm{f}-2\overline{i}+3\overline{j}+5\overline{k}$ and

$7\overline{i}-\overline{k}$ is

0%
$-3:2$
0%
$1:2$
0%
$2:3$
0%
$-4:3$

Explanation

Given

$i+j+k$ divides

$-2\overline{ i } +3\overline { j } +5\overline { k }, 7\overline { i } -\overline { k }$

${ x }_{ 1 }=-2, { x }_{ 2 }=1, { x }_{ 3 }=7$

Now, Required Ratio

$= { x }_{ 1 }-{ x }_{ 2 }: { x }_{ 2 }-{ x }_{ 3 }$ (or)

${ y }_{ 1 }-{ y }_{ 2 } : { y }_{ 2 }-{ y }_{ 3 } =-2-(1) : 1-7 = -3: -6 = 1:2$

OPTION B

Orthocentre of an equilateral triangle

$ABC$ is the origin

$\mathrm{O}$ . If

$A=\overline{a},\ B=\overline{b},\ C=\overline{c}$ then

$\overline{AB}+2\overline{BC}+3\overline{CA}=$

0%
$3\overline{c}$
0%
$3\overline{a}$
0%
$0$
0%
$3\overline{b}$

Explanation

$\overline{AB}+2\overline{BC}+3\overline{CA}$

$=\vec{b}-\vec{a}+2(\vec{c}-\vec{b})+3(\vec{a}-\vec{c})$

$=-\vec{b}+2\vec{a}- \vec{c}$ ----------(1)

orthocentre = circumcenter (property of equalatoral

$\Delta$ )
so circumecentre

$= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$

$0= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$

$-\vec{b}=\vec{a}+\vec{c}$
so put in (1)

$3\vec{a}$

If the vertices of a

$\Delta ABC$ are

$A= (1,-1,-3)$ ,

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

$C=(-5,2,-6)$ then the length of the internal bisector of angle

$A$ is

0%
$\displaystyle \dfrac{3\sqrt{10}}{2}$
0%
$\displaystyle \dfrac{3\sqrt{10}}{5}$
0%
$\displaystyle \dfrac{3\sqrt{10}}{7}$
0%
$\displaystyle \dfrac{3\sqrt{10}}{4}$

Explanation

$AD$ bisects the

$BC$ in ratio of

$\dfrac{|AB|}{|AC|}=\dfrac{|BD|}{|CD|}=\dfrac{\sqrt6}{\sqrt{54}}=\dfrac{1}{3}$

$\dfrac{BD}{DC}=\dfrac{1}{3}=\dfrac{c}{b}$

Coordinates of

$D$ is

$=\dfrac{1}{4}(3B+C)$ (section formula)

$D(\dfrac{1}{4},\dfrac{5}{4},-3)$

Length of

$AD=\sqrt{(1-\dfrac{1}{4})^{2}+(-1-\dfrac{5}{4})^{2}+(-3+3)^{2}}$

$=\dfrac{\sqrt{90}}{4}$

$=\dfrac{3\sqrt{10}}{4}$

The condition for the vectors

$\vec{a},\vec{b},\vec{c},\vec{d}$ to be the sides of a parallelogram taken in order is

0%
$\vec{a}+\vec{b}=\vec{c}+\vec{d}$
0%
$\vec{a}+\vec{b}=\vec{c}+\vec{d}=0$
0%
$\vec{a}+\vec{c}=\vec{b}+\vec{d}$
0%
$\vec{a}+\vec{c}=\vec{b}+\vec{d}=0$

Explanation

Consider the vectors

$\vec{a}$ and

$\vec{b}$ .

Now, the sides with length as

$\vec a , \vec b$ forms two sides of a triangle and diagonal forms the third side of the same triangle.

$\therefore$ By using triangle law of vector addition, the diagonal is given by

$\vec{a}+\vec{b}$ ........... (i)
Now consider the vectors

$\vec{c}$ and

$\vec{d}$ .
The same diagonal can be written as

$\vec{c}+\vec{d}$ using triangle law of vector addition.

$\therefore$

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

Hence, option A is correct.

Let

$A$ and

$B$ be points with position vectors

$\overline{a}$ and

$\overline{b}$ with respect to origin

$O$ . If the point

$C$ on

$OA$ is such that

$2\overline{AC}=\overline{CO}$ ,

$\overline{CD}$ is parallel to

$\overline{OB}$ and

$|\overline{CD}|=3|\overline{OB}|$ then

$AD$ is

0%
$\displaystyle \overline{b}-\dfrac{a}{9}$
0%
$3\displaystyle \overline{b}-\dfrac{a}{3}$
0%
$\displaystyle \overline{b}-\dfrac{a}{3}$
0%
$\displaystyle \overline{b}+\dfrac{a}{3}$

Explanation

Given that

$2 \overrightarrow{AC}=\overrightarrow{CO}$

$2(\vec{c}-\vec{a})=-\vec{c}$

$3\vec{c}=2\vec{a}$

$\vec{c}=\dfrac{2\vec{a}}{3}$

Also given

$\overrightarrow{CD}\left | \right |\overrightarrow{OB}$

$\Rightarrow \vec{a}-\vec{c}=k\vec{b}$

$\Rightarrow \left | \vec{d}-\vec{c} \right |=3\left | \vec{b} \right |$

$\Rightarrow k\left | \vec{b} \right |=3\left | \vec{b} \right |$

$\Rightarrow k=3$
and

$\vec{d}=\vec{c}+3\vec{b}$

$\Rightarrow \overrightarrow{AD}=\vec{d}-\vec{a}$

$=3\vec{b}+\vec{c}-\vec{a}$

$\Rightarrow \overrightarrow{AD}=3\vec{b}-\dfrac{\vec{a}}{3}$

The adjacent sides of a parallelogram are

$2{ \hat { i } }+4{ \hat { j } }-5{ \hat { k } }$ and

${ \hat { i } }+2{ \hat { j } }{ + }3{ \hat { k } }$ then the unit vector parallel to a diagonal is

0%
$\displaystyle \dfrac{-\overline{i}-2\overline{j}+8\overline{k}}{\sqrt{69}}$
0%
$\displaystyle \dfrac{3\overline{i}+6\overline{j}-2\overline{k}}{7}$
0%
$\displaystyle \dfrac{\overline{i}+2\overline{j}-8\overline{k}}{\sqrt{69}}$
0%
All the above

Explanation

As in this question direction of vector is not specified then it can be

$\pm \vec{a}\ \pm \vec{b}=$ Diagonal Vectors
Diagonal vector =

$\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\dfrac{3i+6\hat{j}-2\hat{k}}{\sqrt{49}}$
or

$\dfrac{\vec{a}-\vec{b}}{|\vec{a}-\vec{b}|}=\dfrac{i+2\hat{j}-8\hat{k}}{\sqrt{69}}$

$\dfrac{\vec{-a}+\vec{b}}{|\vec{-a}+\vec{b}|}=\dfrac{-i-2\hat{j}+8\hat{k}}{\sqrt{69}}$

If the diagonals of a parallelogram are

$\overline{i}+5\overline{j}-2\overline{k}$ and

$-2\overline{i}+\overline{j}+3\overline{k}$ , then the lengths of its sides are

0%
$\displaystyle \dfrac{\sqrt{38}}{2},\ \displaystyle \dfrac{\sqrt{50}}{2}$
0%
$\displaystyle \dfrac{\sqrt{35}}{2},\dfrac{\sqrt{45}}{2}$
0%
$\sqrt{38},\ \sqrt{50}$
0%
$\sqrt{35},\ \sqrt{45}$

Explanation

Let

$A$ be origin

$\left | \overrightarrow{AC} \right |= (\hat{i}+5\hat{j}-2\hat{k})$

$(C=1,5,-2)$

$O$ is mid point of

$AC$ ,

$O= (\frac{1}{2}, \frac{5}{2}, -1)$

$B= (a,b,c)$

$\overrightarrow{OB}= \dfrac{1}{2}(\overrightarrow{BD})$

$(a\dfrac{-1}{2})\hat{i}+(b\dfrac{-5}{2})\hat{j}+(c+1)\hat{k}= -\hat{i}+\dfrac{1}{2}\hat{j}+\dfrac{3}{2}\hat{k}$

$a=\dfrac{-1}{2}\ b=3\ c= \dfrac{1}{2}$

$\left | \overrightarrow{AB} \right |= \sqrt{\frac{1}{4}+9+\dfrac{1}{4}}= \sqrt{\dfrac{38}{2}}$

$\left | \overrightarrow{BC} \right |= \sqrt{\frac{9}{4}+4+\dfrac{25}{4}}= \sqrt{\dfrac{50}{2}}$

Let

$A=2\hat{i}+4\hat{j}-\hat{k}$ ,

$B=4\hat{i}+5\hat{j}{+}\hat{k}$ . If the centroid

$G$ of the triangle

$ABC$ is

$3\hat{i}+5\hat{j}-\hat{k}$ , then the position vector of

$C$ is

0%
$3\hat{i}-6\hat{j}{+}3\hat{k}$
0%
$3\hat{i}-6\hat{j}-3\hat{k}$
0%
$3\hat{i}-6\hat{j}{+}2\hat{k}$
0%
$3\hat{i}+6\hat{j}-3\hat{k}$

Explanation

Centroid is given by

$G= \dfrac{\vec{A}+\vec{B}+\vec{C}}{3}$

Putting given values to find

$\vec{C}$

$\Rightarrow \displaystyle 3(3\hat{i}+5\hat{j}-k)= 2\hat{i}+4\hat{j}-\hat{k}+4\hat{i}+5\hat{j}+\hat{k}+\vec{C}$

$\Rightarrow \displaystyle 3\hat{i}+6\hat{j}-3\hat{k}=\vec{C}$

$I$ is the centre of a circle inscribed in a triangle

$ABC$ , then

$|\overline{BC}|\overline{IA}+|\overline{CA}|\overline{IB}+|\overline{AB}|\overline{IC}$ is

0%
$\overline{0}$
0%
$\overline{IA}+\overline{IB}+\overline{IC}$
0%
$\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{3}$
0%
$\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{2}$

Explanation

We know that the position vector of the incenter of a

$\triangle ABC$ is given by

$\displaystyle \frac { BC.\overline { a } +CA.\overline { b } +AB.\overline { c } }{ BC+CA+AB }$

where

$\displaystyle \overline { a } ,\overline { b } ,\overline { c }$ are the affixes of the vertices of the triangle.

Therefore, in the given triangle

$ABC,$ the affix of

$I$ is given by

$\displaystyle \dfrac { \left| \overline { BC } \right| \overline { IA } +\left| \overline { CA } \right| \overline { IB } +\left| \overline { AB } \right| \overline { IC } }{ \left| \overline { BC } \right| +\left| \overline { CA } \right| +\left| \overline { AB } \right| }$

But it is the position vector of

$I$ with respect to

$O.$

$\displaystyle \therefore \ \overline { O } =\dfrac { \left| \overline { BC } \right| \overline { IA } +\left| \overline { CA } \right| \overline { IB } +\left| \overline { AB } \right| \overline { IC } }{ \left| \overline { BC } \right| +\left| \overline { CA } \right| +\left| \overline { AB } \right| }$

$\displaystyle \Rightarrow \left| \overline { BC } \right| \overline { IA } +\left| \overline { CA } \right| \overline { IB } +\left| \overline { AB } \right| \overline { IC } =0$

$\vec{{r}} {\times} \vec{{a}}=\vec{{b}}{\times}\vec{{a}};\ \vec{{r}}{\times}\vec{{b}}=\vec{{a}}{\times}\vec{{b}};\ \vec{a}\neq 0,\vec{b}\neq 0,\vec{a}\neq\lambda\vec{b};\ \vec{a}$ is not perpendicular to

$\vec{b}$ , then

$\vec{r}=$

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

Explanation

Given,

$\vec{r} \times \vec{a} = -(\vec{a} \times \vec{b})$ and

$\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$
Thus,

$\vec{r} \times \vec{a} = -(\vec{r} \times \vec{b})$

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

$\Rightarrow \vec{r} = \lambda(\vec{a} + \vec{b})$

Hence, option B.

Let

$A({\vec{a}})$ ,

$B({\vec{b}}), C({\vec{c}})$ be the vertices of the triangle

$ABC$ and let

$DEF$ be the mid points of the sides

$BC, CA, AB$ respectively. If

$P$ divides the median

$AD$ in the ratio

$2:1$ then the position vector of

$P$ is

0%
$0$
0%
${\vec{a}}+{\vec{b}}+{\vec{c}}$
0%
$\displaystyle \dfrac{\vec{{a}}+\vec{{b}}+\vec{{c}}}{3}$
0%
$\displaystyle \dfrac{2{\vec{a}}+{\vec{b}}+{\vec{c}}}{3}$

Explanation

Given :

$A(\vec{a}), B(\vec{b}), C(\vec{c})$ are the vertices of the triangle

$ABC$ .

$D$ is the midpoint of

$BC$

$\implies D=\left(\dfrac{\vec{b}+\vec{c}}{2}\right)$

$E$ is the midpoint of

$CA$

$\implies E=\left(\dfrac{\vec{a}+\vec{c}}{2}\right)$

$F$ is the midpoint of

$AB$

$\implies F=\left(\dfrac{\vec{a}+\vec{b}}{2}\right)$

Now,

$AD, BE$ and

$CF$ are the medians of triangle

$ABC$

All three medians intersects at point

$P$ .

$\because\ P$ divides

$AD$ in the ratio

$2:1$ and other medians also intersect at

$P$ .

$\implies P$ divides all three medians in the ratio

$2:1$

$\implies P$ is the centroid of the triangle

$ABC$ .

$\implies P= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$ ...... By def

$^n$ of centroid.

Hence, option C is correct.

The plane

$2x-3y+z+6=0$ divides the line segment joining

$(2, 4, 16)$ and

$(3, 5, -4)$ in the ratio

0%
$4 :5$
0%
$4:7$
0%
$2:1$
0%
$1:2$

Explanation

Let the line segment joining points

$P(2,4,16)$ and

$Q(3,5,-4)$ be divided by the given plane in the ratio

$k:1$ at the point

$R.$

Then co-ordinates of

$R$ are

$\displaystyle \left( \frac { 3k+2 }{ k+1 } ,\frac { 5k+4 }{ k+1 } ,\frac { -4k+16 }{ k+1 } \right)$

Since

$R$ lies on the plane

$2x-3y+z+6=0,$

We have,

$\displaystyle 2\left( \frac { 3k+2 }{ k+1 } \right) -3\left( \frac { 5k+4 }{ k+1 } \right) +\left( \frac { -4k+16 }{ k+1 } \right) +6=0$

$\displaystyle \Rightarrow 2\left( 3k+2 \right) -3\left( 5k+4 \right) +\left( -4k+16 \right) +6\left( k+1 \right) =0$

$\displaystyle \Rightarrow -7k+14=0\Rightarrow k=2$

$\therefore PQ$ is divided by the given plane in the ratio

$k:1$ i.e.,

$2:1$

The position vectors of points

$\vec{A},\vec{B},\vec{C}$ are respectively

$\vec{a},\vec{b},\vec{c}$ . If

$P$ divides

$\vec{AB}$ in the ratio

$3:4$ and

$Q$ divides

$\vec{BC}$ in the ratio

$2:1$ both externally then

$\vec{PQ}$ is

0%
$\vec{b}+\vec{c}-\vec{2a}$
0%
$2(\vec{b}+\vec{c}-\vec{2a})$
0%
$4\vec{a}-\vec{b}-\vec{c}$
0%
$\dfrac{-2\vec{a}-\vec{b}-\vec{c}}{2}$

Explanation

By section formula

$\vec p = \dfrac{3 \vec b - 4 \vec a}{3 - 4} = \dfrac{3 \vec b - 4 \vec a}{-1}$ and

$\vec q = \dfrac{2 \vec c - \vec b}{2 - 1} = \dfrac{2 \vec c - \vec b}{1}$

Thus,

$\vec {PQ} = \vec q - \vec p$

$\therefore$

$\vec {PQ} = 2 \vec c - \vec b - (- 3 \vec b + 4 \vec a)$

$\therefore$

$\vec {PQ} = 2 \vec c - \vec b + 3 \vec b - 4 \vec a$

$\therefore$

$\vec {PQ} = 2 \vec c + 2 \vec b - 4 \vec a$

If the position vectors of

$A,B,C$ are

$3\hat{i}+\hat{j}-\hat{k}$ ,

$\hat{i}-2\hat{j}$ ,

$2\hat{i}-\hat{j}{+}3\hat{k}$ then the position vector of the centroid of the triangle

$ABC$ is

0%
$2\hat{i}\ + \dfrac{-2}{3}\hat{j}\ +\dfrac{2}{3}\hat{k}$
0%
$2\hat{i}\ + \dfrac{2}{3}\hat{j}\ +\dfrac{2}{3}\hat{k}$
0%
$3\hat{i}\ + \dfrac{2}{3}\hat{j}\ +\dfrac{4}{3}\hat{k}$
0%
None of these

Explanation

Centroid of a triangle is given by

$G= \dfrac{\vec{A}+\vec{B}+\vec{C}}{3}$
Putting the given values we get

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

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

Let

$\vec{A}=2\hat{i}+7\hat{j},\vec{B}=\hat{i}+2\hat{j}+4\hat{k},\ \displaystyle \vec{C}=\dfrac{9\hat{i}+30\hat{j}+4\hat{k}}{5}$ .

The ratio in which

$\vec{C}$ divides

$\vec{AB}$ internally is?

0%
$1:4$
0%
$2:3$
0%
$3:2$
0%
$5:1$

Explanation

As shown in the figure let

$\vec{C}$ divide

$\vec{AB}$ in a ratio

$K:1$

$C= \dfrac{KB+A}{K+1}$

$\implies \dfrac{9\hat{i}+30\hat{j}+4\hat{k}}{5} = \dfrac{K(\hat{i}+2\hat{j}+4\hat{k})+(2\hat{i}+7\hat{j})}{K+1}$

$\implies 9K\hat{i}+30K\hat{j}+4K\hat{k}+9\hat{i}+30\hat{j}+4\hat{k} = {5K\hat{i}+10K\hat{j}+20K\hat{k}+10\hat{i}+35\hat{j}}$

$\implies 4K\hat{i}+20K\hat{j}-16K\hat{k}=\hat{i}+5\hat{j}-4\hat{k}$

$\implies 4K(\hat{i}+5\hat{j}-4\hat{k})=\hat{i}+5\hat{j}-4\hat{k}$

$\implies K=\dfrac{1}{4}$
Hence,

$\vec{C}$ divide

$\vec{AB}$ in a ratio

$1:4$

$\vec a, \vec b, \vec c, \vec d$ are the position vectors of the points

$A, B, C, D$ respectively such that

$3\vec{a}+5\vec{b}-3\vec{c}-5\vec{d}=0$ then

$AB$ intersects

$CD$ in the ratio

0%
$2:3$
0%
$3:2$
0%
$3:5$
0%
$5:3$

Explanation

Solving the equation

$3 \overline a + 5 \overline b = 3 \overline c + 5 \overline d$

$\therefore$

$\dfrac{3 \overline a + 5 \overline b}{3 + 5} = \dfrac{3 \overline c + 5 \overline d}{3 + 5}$
We see that there is a point such that it's dividing

$AB$ as well as

$CD$ in the above format, thus its equivalent to say that

$AB$ divides

$CD$ in the ratio

$5 : 3$ internally.

$(2, -1, 2)$ is the centroid of tetrahedron

$OABC$ and

$G_{1}$ is the centroid of

$\Delta ABC$ then

$|\overline{OG}_{1}|=$

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$4$
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$1$
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$\dfrac{9}{2}$
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$\dfrac{3}{2}$

Explanation

Centroid of

$OABC= \dfrac{\vec{a}+\vec{b}+\vec{c}+\vec{o}}{4}$

$4\times (2,-1,2)= \vec{a}+\vec{b}+\vec{c}$
Centroid of

$\Delta ABC= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$

$G_{1}= \dfrac{4}{3}(2,-1,2)$

$\left | OG_{1} \right |= \dfrac{4}{3}\sqrt{2^{2}+1^{2}+2^{2}}$

$= 4$

$\overline{a}=\overline{i}+\overline{j}+\overline{k},\overline{b}=2\overline{i}-3\overline{j}+\overline{k}$ , then

$\displaystyle \dfrac{\overline{a}\times\overline{b}}{|\overline{a}\times\overline{b}|}+\dfrac{\overline{b}\times\overline{a}}{|\overline{b}\times\overline{a}|}=$

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$\overline{0}$
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$2\overline{i}+\overline{j}-2\overline{k}$
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$\overline{i}+\overline{j}+2\overline{k}$
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$\overline{i}+2\overline{j}-\overline{k}$

Explanation

$\overline{a} \times \overline{b} = -(\overline{b} \times \overline{a})$
Also,

$|\overline{a} \times \overline{b}| = |(\overline{b} \times \overline{a})|$
Thus, the denominators become equal and the numerator adds up to

$0$ .
Hence the answer becomes

$0$ .

The position vectors of

$A, B , C, D$ are

$\overline{a},\overline{b},\ \overline{c},\overline{d}$ respectively and

$|\overline{a}-\overline{d}|=|\overline{b}-\overline{d}|=|\overline{c}-\overline{d}|$ , then for the triangle

$ABC$ ,

$D$ is

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Ortho centre
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Centroid
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Circumcentre
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Incentre

Explanation

$|\overline{a}-\overline{d}|=|\overline{b}-\overline{d}|=|\overline{c}-\overline{d}|$

$\Rightarrow$ Point

$D$ is equidistant from the points

$A,B,C$ or the vertices of triangle.

$\Rightarrow D$ must be the circumcentre of the triangle.

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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