Vector Algebra Test

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Vector Algebra Test - 39

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Question 1
1 / -0

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 $$\lambda=$$

A
$$\displaystyle \frac { 1 }{ 4 } $$

B
$$\displaystyle \frac { 1 }{ 2 } $$

C
$$\displaystyle \frac { 1 }{ 6 } $$

D
$$\displaystyle \frac { 1 }{ 8 } $$

Solution

$$OM_1=OA+OB+OC+OD$$ (given)
$$=OM+MA+OM+MB+OM+MC+OM+MD$$
$$=4OM+(MA+MC)+(MB+MD)=4OM$$ $$(\because MA=-MC,MB=-MD)$$
$$\displaystyle\therefore OM=\frac{1}{4}OM_1,$$
$$\displaystyle \therefore \lambda=\frac{1}{4}$$
Question 2
1 / -0

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

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

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

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

D
None of these

Solution

$$\overrightarrow { a } \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) =\left( \overrightarrow { a } \cdot \overrightarrow { c } \right) \overrightarrow { b } -\left( \overrightarrow { a } \cdot \overrightarrow { b } \right) \overrightarrow { c } =0-0=0\\ \left( \because \overrightarrow { a } \bot \overrightarrow { b } ,\overrightarrow { a } \bot \overrightarrow { c } ,\therefore \overrightarrow { a } \cdot \overrightarrow { b } =0,\overrightarrow { a } \cdot \overrightarrow { c } =0 \right) $$
Question 3
1 / -0

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 $$\displaystyle \overrightarrow{OS}$$/

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

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

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

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

Solution

P,Q,R,S are the mid-points of OA,OB,BC,AC
$$\overrightarrow { P } =\dfrac { \overrightarrow { A }-\vec{O} }{ 2 } =\dfrac { \overrightarrow { a } }{ 2 } $$
$$\vec{OP}=\vec{P}-\vec{O}$$
$$\vec{OP}=\dfrac{a}{2}-0$$
$$\vec{OP}=\dfrac{a}{2}$$
$$\overrightarrow { Q } =\dfrac { \overrightarrow { b } }{ 2 } $$
$$\vec{OQ}=\vec{Q}-\vec{O}$$
$$\vec{OP}=\dfrac{b}{2}-0$$
$$\vec{OP}=\dfrac{b}{2}$$

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

$$\overrightarrow { S } =\dfrac { \overrightarrow { a } +\overrightarrow { c } }{ 2 }$$
$$\vec{OS}=\vec{S}-\vec{O}$$
$$\vec{OS}=\dfrac{a+c}{2}-0$$
$$\vec{OS}=\dfrac{a+c}{2}$$
Question 4
1 / -0

Given the vectors $$\bar a$$ and $$\bar b$$ as follows. Find the projections of $$\bar a$$ on $$\bar b$$ and of $$\bar b$$ on $$\bar a$$.
$$\bar a=\hat i+\hat j+\hat k$$; $$\displaystyle \bar b= \sqrt{3}\hat i+3\hat j-2\hat k$$

A
$$\displaystyle \frac{\sqrt{2}+1}{5},\frac{3+\sqrt{3}}{5}.$$

B
$$\displaystyle \frac{\sqrt{3}+1}{4},\frac{3+\sqrt{3}}{3}.$$

C
$$\displaystyle \frac{\sqrt{5}+2}{4},\frac{4+\sqrt{3}}{4}.$$

D
$$\displaystyle \frac{\sqrt{2}+5}{3},\frac{4+\sqrt{3}}{3}.$$

Solution

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 $$B$$.
Question 5
1 / -0

In a $$\triangle OAB$$, $$L$$ is a point on the side $$AB$$ and $$M$$ is a point on the side $$OB$$ and the lines $$OL$$ and $$AM$$ meet at $$S$$. It is given that $$AS=SM$$ and $$4 OS=3OL$$. and that $$\displaystyle \left (\frac{OM}{OB} \right )=h$$ and $$\displaystyle \left (\frac{AL}{AB} \right )=k.$$ Express the vectors $$\displaystyle \overrightarrow{AM},$$ and $$\displaystyle \overrightarrow{OS}$$ in term of $$a, b$$ and $$h$$ and the vectors $$\displaystyle \overrightarrow{OL}$$ and $$\displaystyle \overrightarrow{OS}$$ in terms of $$a,b$$ and $$k$$ where $$\displaystyle \overrightarrow{OA}=a$$ and $$\displaystyle \overrightarrow{OB}=b.$$ Find $$h$$ and $$k$$

A
$$ \dfrac{1}{2},\dfrac{1}{3}$$

B
$$ \dfrac{1}{3},\dfrac{1}{2}$$

C
$$ \dfrac{1}{6},\dfrac{1}{2}$$

D
$$ \dfrac{1}{3},\dfrac{1}{6}$$

Solution

Given
$$\vec{OA}=\vec{a},\vec{OC}=\vec{b}$$
Let
$$(\dfrac{OM}{MB})=\dfrac{\lambda}{1}$$
$$(\dfrac{AL}{LB})=\dfrac{\mu}{1}$$
SO point M
$$\vec{OM}=\dfrac{\lambda b}{\lambda+1}$$
and point L
$$\vec{OL}=\dfrac{\mu b+a}{\mu+1}$$

Here $$AS=SM$$ i.e. S is mid point Of $$\vec{OA}$$ and $$\vec{OM}$$
$$\vec{OS}=\dfrac{\vec{a}+\dfrac{\lambda b}{\lambda+1}}{2}$$
$$\vec{OS}=\dfrac{\vec{a}(\lambda+1)+\lambda b}{2(\lambda +1)}$$
$$\vec{OS}=\dfrac{\vec{a}\lambda +\vec{a}+\lambda \vec{b}}{2(\lambda +1)}$$--------(1)

Here $$4OS=3OL$$ i.e $$\dfrac{OS}{SL}=\dfrac{3}{1}$$
$$\vec{OS}=\dfrac{3\vec{OL}}{4}$$
$$\vec{OS}=\dfrac{3\mu b+3a}{4(\mu+1)}$$-------(2)
eq(1)=eq(2)
$$\dfrac{\vec{a}\lambda +\vec{a}+\lambda \vec{b}}{2(\lambda +1)}=\dfrac{3\mu b+3a}{4(\mu+1)}$$
$$\dfrac{\vec{a}\lambda +\vec{a}+\lambda \vec{b}}{(\lambda +1)}=\dfrac{3\mu b+3a}{2(\mu+1)}$$
$$(\vec{a}\lambda +\vec{a}+\lambda \vec{b})(2(\mu+1))=(\lambda +1)(3\mu b+3a)$$
$$2\mu\lambda\vec{a} +2\vec{a}\lambda+2\mu\vec{a}+2\vec{a}+2\mu\lambda\vec{b}+2\lambda\vec{b}=3\lambda\mu \vec{b}+3\mu \vec{b}+3\vec{a}\lambda+3\vec{a}$$
$$(2\mu\lambda+2\lambda+2\mu+2)\vec{a}+(2\mu\lambda+2\lambda)\vec{b}=(3\lambda\mu+3\mu)\vec{b}+(3\lambda+3)\vec{a}$$
comparing both sides w.r to a
$$2\mu\lambda+2\lambda+2\mu+2=3\lambda+3$$
$$2\mu\lambda+2\mu=\lambda+1$$
$$\mu=\dfrac{\lambda+1}{2\lambda+2}$$
$$\mu=\dfrac{1}{2}$$
and on comparing both sides w.r to b
$$2\mu\lambda+2\lambda=3\mu\lambda+3\mu$$
$$2\lambda=\mu\lambda+3\mu$$
putting $$\mu$$ in above eq
$$2\lambda=\dfrac{1}{2}\lambda+\dfrac{3}{2}$$
$$2\lambda-\dfrac{1}{2}\lambda=\dfrac{3}{2}$$
$$\dfrac{3}{2}\lambda=\dfrac{3}{2}$$
$$\lambda=1$$

$$h=(\dfrac{OM}{OB})=\dfrac{\lambda}{2}$$
$$h=(\dfrac{OM}{OB})=\dfrac{1}{2}$$
$$k=(\dfrac{AL}{AB})=\dfrac{AL}{AL+LB}$$
$$k=(\dfrac{AL}{AB})=\dfrac{\mu}{\mu+1}$$
$$k=(\dfrac{AL}{AB})=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+1}$$
$$k=\dfrac{\dfrac{1}{2}}{\dfrac{3}{2}}$$
$$k=\dfrac{1}{3}$$
Question 6
1 / -0

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

A
3/7

B
7/3

C
3

D
7

Solution

$$\displaystyle \lambda =\dfrac{the\:projection\:of\:a\:on\:b}{the\:projection\:of\:b\:on\:a},$$
$$\displaystyle =\dfrac { \dfrac { a.b }{ \left| b \right| } }{ \dfrac { b.a }{ \left| a \right| } } =\dfrac { \left| a \right| }{ \left| b \right| } =\dfrac { 7 }{ 3 } $$
Question 7
1 / -0

Directions For Questions

ABC is a triangle and P,Q are the mid-points of AB, AC respectively. If $$\displaystyle \overrightarrow{AB}=2\vec a$$ and $$\displaystyle \overrightarrow{AC}=2\vec b$$.

...view full instructions

$$\displaystyle \overrightarrow{BC}$$

A
$$\displaystyle \overrightarrow{BC}=2(b-a)$$

B
$$\displaystyle \overrightarrow{BC}=2(a-b)$$

C
$$\displaystyle \overrightarrow{BC}=2(b+a)$$

D
none of these

Solution

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)
$$\vec{BC}=2\vec{b}+\vec{OA}-(\vec{OA}+2\vec{a})$$
$$\vec{BC}=2\vec{b}+\vec{OA}-\vec{OA}-2\vec{a}$$
$$\vec{BC}=2\vec{b}-2\vec{a}$$
$$\vec{BC}=2(\vec{b}-\vec{a})$$
Question 8
1 / -0

If $$\displaystyle \overrightarrow{PO}+\overrightarrow{OQ}=\overrightarrow{QO}+\overrightarrow{OR},$$ then $$\displaystyle P, Q, R$$ are

A
the vertices of an equilateral triangle

B
the vertices of an isosceles triangle

C
collinear

D
None of these

Solution

$$\displaystyle \overrightarrow{PO}+\overrightarrow{OQ}=\overrightarrow{QO}+\overrightarrow{OR}$$
$$\Rightarrow\displaystyle 2\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{OR}$$
$$\Rightarrow \displaystyle\overrightarrow{OQ}=\dfrac{\overrightarrow{OP}+\overrightarrow{OR}}{2}$$
$$\therefore P,Q,R$$ are collinear.
Hence, option C.
Question 9
1 / -0

Projection of the vector $$2i + 3j - 2k$$ on the vector $$i - 2j + 3k$$ is

A
$$\displaystyle 2/\sqrt{\left ( 14 \right )}$$

B
$$\displaystyle 1/\sqrt{\left ( 14 \right )}$$

C
$$\displaystyle 3/\sqrt{\left ( 14 \right )}$$

D
None of these

Solution

Projection of $$a$$ and $$b$$ is
$$\displaystyle \dfrac { a\cdot b }{ \left| b \right| } =\dfrac { \left( 2i+3j-2k \right) \cdot \left( i-2j+3k \right) }{ \left| i-2j+3k \right| } =\dfrac { 2 }{ \sqrt { 14 } } $$
Question 10
1 / -0

Directions For Questions

$$ABCD$$ is parallelogram whose diagonals intersect at $$E$$ and $$M$$ is the mid-Point of $$DC$$. If$$\displaystyle \overrightarrow{AB}=\bar a$$ and $$\displaystyle \overrightarrow{AD}=\bar b,$$express in terms of $$\bar a$$ and $$\bar b$$ the vectors

...view full instructions

$$\displaystyle \overrightarrow{AE}$$

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

B
$$\displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar a-\bar b \right )$$

C
$$\displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar b-\bar a \right )$$

D
none of these

Solution

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.

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Vector Algebra Test - 39

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Vector Algebra Test - 39

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SHARING IS CARING

Question 1

$$\displaystyle \frac { 1 }{ 4 } $$

$$\displaystyle \frac { 1 }{ 2 } $$

$$\displaystyle \frac { 1 }{ 6 } $$

$$\displaystyle \frac { 1 }{ 8 } $$

Solution

Question 2

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

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

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

None of these

Solution

Question 3

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

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

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

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

Solution

Question 4

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

$$\displaystyle \frac{\sqrt{3}+1}{4},\frac{3+\sqrt{3}}{3}.$$

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

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

Solution

Question 5

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

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

$$ \dfrac{1}{6},\dfrac{1}{2}$$

$$ \dfrac{1}{3},\dfrac{1}{6}$$

Solution

Question 6

3/7

7/3

3

7

Solution

Question 7

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

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

$$\displaystyle \overrightarrow{BC}=2(b+a)$$

none of these

Solution

Question 8

the vertices of an equilateral triangle

the vertices of an isosceles triangle

collinear

None of these

Solution

Question 9

$$\displaystyle 2/\sqrt{\left ( 14 \right )}$$

$$\displaystyle 1/\sqrt{\left ( 14 \right )}$$

$$\displaystyle 3/\sqrt{\left ( 14 \right )}$$

None of these

Solution

Question 10

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

$$\displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar a-\bar b \right )$$

$$\displaystyle \overrightarrow{AE}=\frac{1}{2}\left ( \bar b-\bar a \right )$$

none of these

Solution

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