Vector Algebra Test

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Vector Algebra Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Let $$A=(-3,4,-8)$$ and $$B=(5,-6,4)$$, then the coordinates of the point in which the $$XY$$- plane or $$XOY$$ plane divides the line segment $$AB$$ is

A
$$(7,-8,0)$$

B
$$(\displaystyle \dfrac{7}{3},-\dfrac{8}{3},0)$$

C
$$(\displaystyle \dfrac{7}{2},-\dfrac{8}{2},0)$$

D
$$(0,\displaystyle \dfrac{7}{3},-\dfrac{8}{3})$$

Solution

Let $$\vec{O}$$ divide $$AB$$ into $$K:1$$
By Section Formula
$$(x,y,o)= \dfrac{KB+A}{K+1}$$
$$(x,y,o)= \dfrac{K(5,-6,4)+(-3,4,-8)}{K+1}$$
$$0= \dfrac{K(4)-8}{K+1}$$
$$K=2$$

$$x=\dfrac{10-3}{3}= \dfrac{7}{3}$$

$$y=\dfrac{-12+4}{3}= \dfrac{-8}{3}$$
Question 2
1 / -0

If $$3\vec{a}+4\vec{b}-7\vec{c}=0$$ then the ratio in which $$C(\vec{c})$$ divides the join of $$A(\vec{a})$$ and $$B(\vec{b})$$ is

A
$$1:2$$

B
$$2:3$$

C
$$3:2$$

D
$$4:3$$

Solution

Use section formula
$$C= \dfrac{K\vec{b}+\vec{a}}{K+1}$$
$$(K+1)\vec{c}= K\vec{b}+\vec{a}$$ ---- (1)
$$a+\dfrac{4}{3}\vec{b}-\dfrac{7}{3}\vec{c}= 0$$ -----(2)
$$\implies K= \dfrac{4}{3}$$
Question 3
1 / -0

The resultant of two concurrent forces $$n\vec{OP}$$ and $$m\vec{OQ}$$ is $$(m+n)\vec{OR}$$. Then $$R$$ divides $$PQ$$ in the ratio

A
$$m:n$$

B
$$n:m$$

C
$$1:n$$

D
$$m:1$$

Solution

Applying Section Formula
$$R= \dfrac{KQ+P}{K+1}$$
$$(K+1)R= KQ+P$$
$${K+1}= \dfrac{m+n}{n}$$
$${K}= \dfrac{m}{n}$$
Question 4
1 / -0

Let $$\vec{A}=\hat{i}+2\hat{j}{+}3\hat{k},\ \vec{B}=4\hat{i}+2\hat{j},\ \vec{C}=2\hat{i}+2\hat{j}{+}2\hat{k}$$. Then the ratio in which $$C$$ divides $$AB$$ is

A
$$3:4$$

B
$$1:3$$

C
$$1:2$$

D
$$1:1$$

Solution

Let $$A=(1,2,3)$$, $$B=(4,2,0)$$ and $$C=(2,2,2)$$
Now let $$C$$ divide the line joining $$BA$$ in the ration $$M:N$$.
Then $$C_{Z}=\dfrac{MA_{Z}+NB_{Z}}{M+N}$$
$$\Rightarrow 2=\dfrac{M\times3+N\times0}{M+N}$$
$$\Rightarrow 2M+2N=3M$$
$$\Rightarrow 2N=M$$
$$\Rightarrow \dfrac{N}{M}=\dfrac{1}{2}$$
Hence, the required ratio is $$1:2$$.
Question 5
1 / -0

$$P, Q, R, S$$ have position vectors $$\overline{p},\overline{q},\overline{r},\overline{s}$$ respectively such that $$\overline{p}-\overline{q}=2(\overline{s}-\overline{r})$$, then which of the following is correct

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

Given : $$P,Q,R,S$$ have position vectors $$\bar{p}, \bar{q}, \bar{r}, \bar{s}$$
$$\bar{p}-\bar{q}=2(\bar{s}-\bar{r})$$
$$\implies \bar{p}-\bar{q}=2\bar{s}-2\bar{r}$$
$$\implies \bar{p}+2\bar{r}=\bar{q}+2\bar{s}$$
$$\implies \dfrac{\vec{p}+2\vec{r}}{3}= \dfrac{\vec{q}+2\vec{s}}{3} $$
$$\implies \dfrac{\vec{P}+2\vec{R}}{3}= \dfrac{\vec{Q}+2\vec{S}}{3} $$
Thus, $$QS$$ & $$PR$$ trisect each other.
Question 6
1 / -0

If $$\vec{a}+2\vec{b},2\vec{a}+\vec{b}$$ be the position vectors of the points $$A$$ and $$B$$, then the position vector of the point $$C$$ which divides $$AB$$ internally in the ratio $$2:1$$ is

A
$$\displaystyle \dfrac{5\vec{a}-4\vec{b}}{3}$$

B
$$\displaystyle \dfrac{5\vec{a}+4\vec{b}}{3}$$

C
$$\displaystyle \dfrac{5\vec{a}-2\vec{b}}{3}$$

D
$$\displaystyle \dfrac{5\vec{a}+2\vec{b}}{3}$$

Solution

Applying Section Formula
$$C= \dfrac{2B+A}{3}$$
$$= \dfrac{4\vec{a}+2\vec{b}+\vec{a}+2\vec{b}}{3}$$
$$C= \dfrac{5\vec{a}+4\vec{b}}{3}$$
Hence, option B is correct.
Question 7
1 / -0

In $$\Delta ABC$$, $$ P,\ Q,\ R$$ are points on $$BC,\ CA,\ AB$$ respectively, dividing them in the ratio $$1:4$$, $$3:2$$ and $$3 : 7$$. The point $$S$$ divides $$AB$$ in the ratio $$1:3$$. Then $$\displaystyle \dfrac{|\overline{AP}+\overline{BQ}+\overline{CR}|}{|\overline{CS}|}$$ is

A
$$\displaystyle \dfrac{1}{5}$$

B
$$\displaystyle \dfrac{2}{5}$$

C
$$\displaystyle \dfrac{5}{2}$$

D
$$\displaystyle \dfrac{7}{10}$$

Solution

P.v of P $$= \dfrac{\vec{c}+4\vec{b}}{5}$$

P.v of R $$= \dfrac{7\vec{a}+3\vec{b}}{10}$$

P.v of Q$$= \dfrac{3\vec{a}+2\vec{b}}{5}$$

P.v of S $$= \dfrac{\vec{b}+3\vec{a}}{4}$$

$$\overrightarrow{AP}= \dfrac{\vec{c}+4\vec{b}-5\vec{a}}{5}\ \ \overrightarrow{BQ}= \dfrac{3\vec{a}+2\vec{c}-5\vec{b}}{5}$$

$$\overrightarrow{CR}= \dfrac{7\vec{a}+3\vec{b}-10\vec{a}}{10}\ \ \overrightarrow{CS}= \dfrac{\vec{b}+3\vec{a}-4\vec{c}}{4}$$

$$\therefore$$ $$\dfrac{\dfrac{\left | 2\vec{c}+4\vec{c}-10\vec{c}+8\vec{b}-10\vec{b}+3\vec{b}-10\vec{a}+6\vec{a}+7\vec{a} \right |}{10}}{\left | \dfrac{\vec{b}+3\vec{a}-4\vec{c}}{4} \right |}$$
$$= \dfrac{4}{10}$$
Question 8
1 / -0

If $$\vec a=\hat i+2\hat j$$ and $$\vec b = 3\hat j$$, then $$\vec a\cdot\vec b=$$

A
3

B
-3

C
6

D
-6

Solution

Yes, a triangular pyramid has 4 triangular faces.
Question 9
1 / -0

If $$\overline{a}$$ and $$\overline{b}$$ are position vectors of $$A$$ and $$B$$ respectively, then the position vector of a point $$C$$ in $$AB$$ produced such that $$\overline{AC}=3\overline{AB}$$ is

A
$$3\overline{a}-\overline{b}$$

B
$$3\overline{b}-\overline{a}$$

C
$$3\overline{a}-2\overline{b}$$

D
$$3\overline{b}-2\overline{a}$$

Solution

Let the Position Vector of $$C$$ be $$\vec{c}$$
We have $$\vec{AC}=3\vec{AB}$$
$$\vec{c} - \vec{a}= 3(\vec{b} - \vec{a})$$
$$\vec{c}= 3\vec{b} - 2\vec{a}$$
Question 10
1 / -0

If $$\left[ \overrightarrow { a } \overrightarrow { b } \overrightarrow { c } \right] =1$$ then $$\frac { \overrightarrow { a } .\overrightarrow { b }\times \overrightarrow { c } }{ \overrightarrow { c }\times \overrightarrow { a } .\overrightarrow { b } } +\frac { \overrightarrow { b } .\overrightarrow { c }\times \overrightarrow { a } }{ \overrightarrow { a }\times \overrightarrow { b } .\overrightarrow { c } } +\frac { \overrightarrow { c } .\overrightarrow { a }\times \overrightarrow { b } }{ \overrightarrow { b }\times \overrightarrow { c } .\overrightarrow { a } }$$ is equal to

A
3

B
1

C
-1

D
None of these

Solution

$$\overrightarrow { a } .\overrightarrow { b } \times \overrightarrow { c } $$ is $$[\overrightarrow { a } \overrightarrow { b } \overrightarrow { c } ]$$
We snow that
$$[\overrightarrow { a } \overrightarrow { b } \overrightarrow { c } ]=[\overrightarrow { c } \overrightarrow { a } \overrightarrow { b } ]$$
Similarly,
$$[\overrightarrow { b } \overrightarrow { c } \overrightarrow { a } ]=[\overrightarrow { a } \overrightarrow { b } \overrightarrow { c } ]$$
= $$[\overrightarrow { c } \overrightarrow { a } \overrightarrow { b } ]$$
$$\therefore $$ Expression = 1 + 1 + 1
= 3