Vector Algebra Test

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Vector Algebra Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

$$P, Q, R, S$$ have position vectors $$p, q, r, s$$, such that $$\displaystyle p-q=2\left ( s-r \right )$$, then

A
$$PQ$$ and $$RS$$ bisect each other

B
$$PQ$$ and $$PR$$ bisect each other

C
$$PQ$$ and $$RS$$ trisect each other

D
$$QS$$ and $$PR$$ trisect each other

Solution

$$\overrightarrow { p } -\overrightarrow { q } =2\left( \overrightarrow { s } -\overrightarrow { r } \right) \Rightarrow \overrightarrow { p } +2\overrightarrow { r } =2\overrightarrow { s } +\overrightarrow { q } =\overrightarrow { q } +2\overrightarrow { s } $$
$$\displaystyle\Rightarrow \dfrac { \overrightarrow { p } +\overrightarrow { 2 } r }{ 1+2 } =\dfrac { \overrightarrow { q } +2\overrightarrow { s } }{ 1+2 } $$
$$\therefore $$ the point divides $$PR$$ in the ratio $$2:1$$, divides $$QS$$ in the ratio $$2:1$$.
$$\therefore $$ $$PR$$ and $$QS$$ trisect each-other
Question 2
1 / -0

OABC is a tetrahedron express the vectors $$\displaystyle \overrightarrow{BC},\overrightarrow{CA},\:and\:\overrightarrow{AB}$$ in terms of the vectors $$\displaystyle \overrightarrow{OA},\overrightarrow{OB}\:and\:\overrightarrow{OC}$$

A
$$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}+\overrightarrow{OB}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}+\overrightarrow{OC}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}+\overrightarrow{OA}$$

B
$$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}-\overrightarrow{OB}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}-\overrightarrow{OC}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}-\overrightarrow{OA}$$

C
$$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}-\overrightarrow{BO}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}-\overrightarrow{CO}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}-\overrightarrow{AO}$$

D
none of these

Solution

$$OABC$$ 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.
Question 3
1 / -0

$$ABC$$ is any triangle and $$D, E, F$$ are the middle points of its sides $$BC, CA, AB$$ respectively. Express the vectors $$\displaystyle \overrightarrow{CF}$$ and $$\overrightarrow{BE}$$ as linear combination of the vectors $$\displaystyle \overrightarrow{AB}$$ and $$\overrightarrow{AC}$$

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

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

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

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

Solution

$$\overrightarrow { CA } +\overrightarrow { AF } =\overrightarrow { CF } $$
$$\Rightarrow -\overrightarrow { AC } +\dfrac { 1 }{ 2 } \overrightarrow { AB } =\overrightarrow { CF } $$
Similarly $$\overrightarrow { AE } +\overrightarrow { BA } =\overrightarrow { BE } $$
$$\Rightarrow \dfrac { 1 }{ 2 } \overrightarrow { AC } -\overrightarrow { AB } =\overrightarrow { BE } $$
Question 4
1 / -0

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

A
$$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } $$

B
$$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 2 } $$

C
$$\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } $$

D
None of these

Solution

$$A\left( \overrightarrow { a } \right) ,B\left( \overrightarrow { b } \right) ,C\left( \overrightarrow { c } \right) $$ ate the vertices of the $$\triangle ABC$$.
Circumcentre is at origin $$\displaystyle G\left( \overrightarrow { g } \right) =\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } $$
$$\displaystyle G=\dfrac { 2\overrightarrow { c } +\overrightarrow { O } }{ 3 } \Rightarrow \overrightarrow { O } =3\overrightarrow { G } $$
$$\displaystyle \Rightarrow \overrightarrow { O } =3\left( \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } \right) =\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } $$
Question 5
1 / -0

If a, b, c are position vectors of the vertices of a $$\displaystyle \Delta ABC,$$ then $$ \displaystyle \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=$$

A
0

B
2a

C
2b

D
3c

Solution

If we join head to tail all the vectors, then we end up at the initial point where we started, that is vertice A.
Hence the net sum is 0.
Question 6
1 / -0

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

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

B
$$\displaystyle \overrightarrow{OC}= 2\vec{b}-2\vec{a}$$

C
$$\displaystyle \overrightarrow{OC}= 2\vec{b}-\vec{a}$$

D
$$\displaystyle \overrightarrow{OC}= 3\vec{a}-2\vec{b}$$

Solution

$$B$$ 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.
Question 7
1 / -0

Given that the vectors $$\bar a,\bar b,\bar c$$ form a base, find the sum of co-ordinates of the vector: $$3\bar u-\bar v+\bar w$$
if $$\bar u=\bar a+\bar c, \bar v=\bar b+\bar c,\bar w=\bar a-\bar b$$;

A
4

B
6

C
2

D
3

Solution

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.
Question 8
1 / -0

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

A
$$\displaystyle \overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{PC}$$

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

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

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

Solution

$$\overrightarrow { AC } $$ 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)
$$\overrightarrow { PA } +\overrightarrow { PB } +\left( \overrightarrow { AC } +\overrightarrow { BC } \right) =2\overrightarrow { PC } \\ \Rightarrow \overrightarrow { PA } +\overrightarrow { PB } =2\overrightarrow { PC } $$
Question 9
1 / -0

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

A
$$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{AD}\quad;\:\overrightarrow{BD}= -\overrightarrow{AB}+\overrightarrow{AD}$$

B
$$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}-\overrightarrow{CB}\quad;\:\overrightarrow{BD}= -\overrightarrow{AB}+\overrightarrow{AD}$$

C
$$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{AD}\quad;\:\overrightarrow{BD}= \overrightarrow{BC}+\overrightarrow{CD}$$

D
all of the above

Solution

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.
Question 10
1 / -0

If $$\displaystyle \left ( \bar A+\bar B \right )$$ is perpendicular to$$\bar B$$ and If $$\displaystyle \left ( \bar A+2\bar B \right )$$ is perpendicular to $$\bar A$$, then

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

B
$$\displaystyle A=2B$$

C
$$\displaystyle 2A=B$$

D
$$\displaystyle A=B$$

Solution

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.

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Vector Algebra Test - 38

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Vector Algebra Test - 38

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SHARING IS CARING

Question 1

$$PQ$$ and $$RS$$ bisect each other

$$PQ$$ and $$PR$$ bisect each other

$$PQ$$ and $$RS$$ trisect each other

$$QS$$ and $$PR$$ trisect each other

Solution

Question 2

$$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}+\overrightarrow{OB}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}+\overrightarrow{OC}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}+\overrightarrow{OA}$$

$$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}-\overrightarrow{OB}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}-\overrightarrow{OC}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}-\overrightarrow{OA}$$

$$\displaystyle \overrightarrow{BC}= \overrightarrow{OC}-\overrightarrow{BO}\quad;\:\overrightarrow{CA}= \overrightarrow{OA}-\overrightarrow{CO}\quad;\:\overrightarrow{AB}= \overrightarrow{OB}-\overrightarrow{AO}$$

none of these

Solution

Question 3

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

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

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

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

Solution

Question 4

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

$$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 2 } $$

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

None of these

Solution

Question 5

0

2a

2b

3c

Solution

Question 6

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

$$\displaystyle \overrightarrow{OC}= 2\vec{b}-2\vec{a}$$

$$\displaystyle \overrightarrow{OC}= 2\vec{b}-\vec{a}$$

$$\displaystyle \overrightarrow{OC}= 3\vec{a}-2\vec{b}$$

Solution

Question 7

4

6

2

3

Solution

Question 8

$$\displaystyle \overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{PC}$$

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

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

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

Solution

Question 9

$$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{AD}\quad;\:\overrightarrow{BD}= -\overrightarrow{AB}+\overrightarrow{AD}$$

$$\displaystyle \overrightarrow{AC}= \overrightarrow{AB}-\overrightarrow{CB}\quad;\:\overrightarrow{BD}= -\overrightarrow{AB}+\overrightarrow{AD}$$

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

all of the above

Solution

Question 10

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

$$\displaystyle A=2B$$

$$\displaystyle 2A=B$$

$$\displaystyle A=B$$

Solution

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