Vector Algebra Test

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Vector Algebra Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

In a triangle $$ABC, D$$ and $$E$$ are points on $$BC$$ and $$AC$$ respectively, such that $$BD=2DC, AE=3EC,$$ Let $$P$$ be the point of intersection of $$AD$$ and $$BE$$. Then $$\dfrac{BE}{PE}=$$

A
$$2:3$$

B
$$3:8$$

C
$$8:3$$

D
$$1:2$$

Solution

Let p.v of $$\vec{a}=\vec{a}$$
p.v of $$\vec{B}=\vec{b}$$
p.v of $$\vec{C}=\vec{c}$$
p.v of $$E=\dfrac{3\vec{c}}{4}$$
p.v of $$D=\dfrac{3\vec{c}}{4}$$
$$=\dfrac{2\vec{c}+\vec{b}}{3}$$
$$p$$ lie on line $$AD$$ so p.v of $$p=b\left ( \dfrac{2\hat{c}+\hat{b}}{3} \right )$$..............(1)
Let $$\dfrac{BP}{PA}=K$$
So p.v of $$p=\dfrac{K\dfrac{3\hat{c}}{4}+\hat{b}}{K+1}$$.............(2)
equate (1) and (2)
$$K=\dfrac{8}{3}$$
Question 2
1 / -0

If $$\overline{a}+\overline{b}+\overline{c}=\alpha\overline{d},\overline{b}+\overline{c}+\overline{d}=\beta\overline{a}$$, then $$\overline{a}+\overline{b}+\overline{c}+\overline{d}$$ is

A
$$(\beta+1)\overline{a}$$

B
$$\alpha\overline{a}$$

C
$$(\alpha+1)\overline{b}$$

D
$$\alpha\overline{a}+\beta\overline{b}$$

Solution
Question 3
1 / -0

If $$OABC$$ is a parallelogram with $$\vec{OB}=\vec{a},\vec{AB}=\vec{b}$$ then $$\vec{OA}=$$

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

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

C
$$\displaystyle \dfrac{1}{2}(\vec{a}+\vec{b})$$

D
$$\displaystyle \dfrac{1}{2}(\vec{a}-\vec{b})$$

Solution

From figure we get
$$\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$$
$$\overrightarrow{OA}=\vec{a}-\vec{b}$$
Question 4
1 / -0

Let us define, the length of a vector $$a\overline{i}+b\overline{j}+c\overline{k}$$ as $$|{a}|+|{b}|+|{c}|$$. This definition coincides with the usual definition of the length of a vector $$a\overline{i}+b\overline{j}+c\overline{k}$$ if

A
$$a=b=c=0$$

B
Any one of $$a, b, c$$, is zero

C
Any two of $$a, b, c$$ are zero

D
$$a=b=c\neq 0$$

Solution

The definitions will coincides if $$|a|+|b|+|c|=\sqrt{a^2+b^2+c^2}$$
$$\Rightarrow a^2+b^2+c^2+2(|a||b|+|b||c|+|c||a|)=a^2+b^2+c^2$$, square both sides
$$\Rightarrow |a||b|+|b||c|+|c||a|=0$$
Clearly above equation will satisfy if any two of $$a,b,c$$ are zero.

Question 5

1 / -0

In $$\Delta ABC,\ D,\ E,\ F$$ are midpoints of the sides $$BC, CA$$ and $$AB$$ respectively. $$O$$' is the circumcentre, $$G$$' is the centroid, $$H$$' is the orthocentre and $$P$$ is any point.
Match the following

List I	List II
$$1) \vec{PA} +\vec{PB}+\vec{PC}$$	$$a) $$$$0$$
$$2)\vec {GA}+\vec{GB}+\vec{GC}$$	$$b) \vec{OH}$$
$$3)\displaystyle \vec{AD}+\dfrac{2}{3}\vec{BE}+\dfrac{1}{3}\vec{CF}$$	$$c)\vec{ PD}+\vec{PE}+\vec{PF}$$
$$4)\vec{OA}+\vec{OB}+\vec{OC}$$	$$d){\displaystyle\dfrac{1}{2}}\vec{AC}$$

$$1-a,2- b,3- c,4- d$$

$$1-c,2- a,3- b,4- c$$

$$1-c,2- a,3- d,4- b$$

$$1-a,2- b,3- d,4- c$$

Solution

1. Since there is an arbitrary point $$P$$ in the question, the answer needs to be dependent on $$P$$ and so, there's only one option which has that, implying the answer to be $$c$$.
2. $$G = \dfrac{a + b + c}{3}$$ where $$a,b,c$$ are the position vectors of $$A,B,C$$ respectively.
Thus, $$\vec {GA} + \vec{ GB} + \vec {GC} = G - a + G - b + G - c = 3G - a - b - c = 0$$
3. We have $$D = \dfrac{b + c}{2}, E = \dfrac{c + a}{2}, F = \dfrac{a + b}{2}$$
Thus, $$\vec{AD} + \dfrac{2}{3}\vec{BE} + \dfrac{1}{3}\vec{CF} = D - a + \dfrac{2}{3}E - \dfrac{2}{3}b + \dfrac{1}{3}F - \dfrac{1}{3}c$$
$$= \dfrac{1}{6}\left[3b + 3c - 6a + 2c + 2a - 4b + a + b - 2c \right] = \dfrac{1}{2}\left[c - a\right]$$, hence option $$d$$.
4. Option $$b$$ is the remaining option!

Question 6
1 / -0

Vector area is a vector quantity associated with each plane figure whose magnitude is

A
Any quantity and direction parallel to the plane

B
Any quantity and direction perpendicular to the plane

C
Equal to the area and direction parallel to the plane

D
Equal to the area and direction perpendicular to the plane

Solution

By definition, the magnitude of vector area of a plane is equal to the area of the plane and has a direction, parallel to the direction of the normal of the plane, i.e. perpendicular to the plane.
Question 7
1 / -0

Let $$OABC$$ be a parallelogram and $$D$$ the midpoint of $$OA$$. The ratio in which $$OB$$ divides $$CD$$ in the ratio

A
$$1:2$$

B
$$1:3$$

C
$$1:4$$

D
$$2:1$$

Solution

P.v. of $$D =\dfrac{\vec{a}}{2}$$
P.v. of $$E=\dfrac{\lambda (\vec{b}+\vec{a})}{}$$
Let $$E$$ divide $$CD$$ in ratio $$K:1$$
P.v. of $$E=\dfrac{K\dfrac{\vec{a}}{2}+\vec{b}}{K+1}$$
$$\lambda (\vec{b}+\vec{a})=\dfrac{\dfrac{k\vec{a}}{2}+\vec{b}}{k+1}$$
$$\lambda =\dfrac{1}{k+1} \lambda =\dfrac{k}{2k+1}$$
So $$ K=2$$
Question 8
1 / -0

In $$\Delta OAB$$, if $$\vec{OA}=\vec{a},\ \vec{OB}=\vec{b}. L$$ is mid point of $$\vec{OA}$$ and $$M$$ is point on $$\vec{OB}$$ such that $$\vec{OM}:\vec{MB}=2:1$$. If $$P$$ is mid point of $$LM$$ then $$\vec{AP}=$$

A
$${\dfrac{1}{3}\vec{b}-\dfrac{3}{4}\vec{a}}$$

B
$$\displaystyle \dfrac{1}{3}\vec{b}+\dfrac{3}{4}\vec{a}$$

C
$${\dfrac{1}{3}\vec{a}-\dfrac{3}{4}\vec{b}}$$

D
$$\displaystyle \dfrac{1}{3}\vec{a}+\dfrac{3}{4}\vec{b}$$

Solution

P.v. of $$ L=\dfrac{\vec{a}}{2}$$ (mid point of $$\vec{OA}$$)
P.v. of $$M=\dfrac{2\vec{b}}{3}$$ (use section formula)
p.v. of $$P=\dfrac{\dfrac{a}{2}+\dfrac{2\vec{b}}{3}}{2}$$ (mid point of LM)
$$\vec{AP}=\dfrac{\vec{a}}{4}+\dfrac{\vec{b}}{3}-\vec{a}$$
$$=\dfrac{-3\vec{a}}{4}+\dfrac{\vec{b}}{3}$$
Question 9
1 / -0

If the vectors $$\vec{c},\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$$ and $$\vec{b}=\hat{j}$$ are such that $$\vec{a},\vec{c}$$ and $$\vec{b}$$ form a right handed system, then $$\vec{c}=$$

A
$$z\hat{i}$$

B
$$y\hat{i}$$

C
$$z\hat{i}-x\hat{k}$$

D
$${-z\hat{i}+x\hat{k}}$$

Solution

$$\vec{b}\times \vec{a}=\vec{c}$$ (as it follow RHS)
$$\vec{b}\times \vec{a}=\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
0 & 1 & 0\\
x & y & z
\end{vmatrix}=z\hat{i}-x\hat{k}=\vec{c}$$
Question 10
1 / -0

The vector $$\vec{AB}=3\hat{i}+4\hat{k}$$ and $$\vec{AC}=5\hat{i}-2\hat{j}+4\hat{k}$$ are the sides of a $$\Delta ABC$$ where $$A$$ is the origin. The length of median through $${A}$$ is

A
$$\sqrt{72}$$

B
$$\sqrt{33}$$

C
$$\sqrt{288}$$

D
$$\sqrt{18}$$

Solution

Let $$A$$ be the origin.
$$D$$ is mid of $$BC=(4,-1,4)$$
So Length $$AD=\sqrt{16+16+1}$$
$$=\sqrt{33}$$

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Vector Algebra Test - 28

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Vector Algebra Test - 28

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Question 1

$$2:3$$

$$3:8$$

$$8:3$$

$$1:2$$

Solution

Question 2

$$(\beta+1)\overline{a}$$

$$\alpha\overline{a}$$

$$(\alpha+1)\overline{b}$$

$$\alpha\overline{a}+\beta\overline{b}$$

Solution

Question 3

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

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

$$\displaystyle \dfrac{1}{2}(\vec{a}+\vec{b})$$

$$\displaystyle \dfrac{1}{2}(\vec{a}-\vec{b})$$

Solution

Question 4

$$a=b=c=0$$

Any one of $$a, b, c$$, is zero

Any two of $$a, b, c$$ are zero

$$a=b=c\neq 0$$

Solution

Question 5

$$1-a,2- b,3- c,4- d$$

$$1-c,2- a,3- b,4- c$$

$$1-c,2- a,3- d,4- b$$

$$1-a,2- b,3- d,4- c$$

Solution

Question 6

Any quantity and direction parallel to the plane

Any quantity and direction perpendicular to the plane

Equal to the area and direction parallel to the plane

Equal to the area and direction perpendicular to the plane

Solution

Question 7

$$1:2$$

$$1:3$$

$$1:4$$

$$2:1$$

Solution

Question 8

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

$$\displaystyle \dfrac{1}{3}\vec{b}+\dfrac{3}{4}\vec{a}$$

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

$$\displaystyle \dfrac{1}{3}\vec{a}+\dfrac{3}{4}\vec{b}$$

Solution

Question 9

$$z\hat{i}$$

$$y\hat{i}$$

$$z\hat{i}-x\hat{k}$$

$${-z\hat{i}+x\hat{k}}$$

Solution

Question 10

$$\sqrt{72}$$

$$\sqrt{33}$$

$$\sqrt{288}$$

$$\sqrt{18}$$

Solution

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