Vector Algebra Test

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Vector Algebra Test - 67

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Weekly Quiz Competition

Question 1
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$$ABCD$$ is a quadrilateral, $$E$$ is the point of intersection of the line joining the midpoints of the opposite sides. If $$O$$ is any point and $$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{x OE},$$ then $$x$$ is equal to

A
$$3$$

B
$$9$$

C
$$7$$

D
$$4$$

Solution

Let $$\vec{OA} = \vec{a}, \vec{OB} = \vec{b}, \vec{OC} = \vec{c}$$ and $$\vec{OD} = \vec{d}$$.

Therefore,
$$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{a} + \vec{b} + \vec{c} + \vec{d}$$.
The midpoint $$P$$ of $$AB$$, is $$\displaystyle \dfrac{\vec{a} + \vec{b}}{2}$$.
The position vector of the midpoint $$Q$$ of $${CD}$$, is $$\displaystyle \dfrac{\vec{c} + \vec{d}}{2}$$.

Therefore, the position vector of the midpoint of $${PQ} $$ is $$\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$$.

Similarly, the position vector of the midpoint of $$RS$$ is $$\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$$, i.e., $$\vec{OE} =\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4} \Rightarrow x = 4$$

Hence, option '$$D$$' is correct.
Question 2
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If $$\overrightarrow{b}$$ is a vector whose initial point divides the join of $$5\widehat{i}$$ and $$5\widehat{j}$$ in the ratio $$k : 1$$ and whose terminal point is the origin and $$|\vec b| \leq \sqrt{37}$$, then $$k$$ lies in the interval

A
$$\left[-6, -\dfrac{1}{6}\right]$$

B
$$\left(- \infty, -6 \right] \cup \left[-\dfrac{1}{6}, \infty \right) $$

C
$$ \left[0, 6 \right] $$

D
None of these

Solution

The point that divides 5$$\widehat{i}$$ and 5$$\widehat{j}$$ in the ratio of $$k : 1$$ is
$$\displaystyle \dfrac{(5 \widehat{j}) k + (5\widehat{i})1}{k + 1}$$
$$\therefore \displaystyle \vec b = \frac{5 \widehat{i} + 5 k \widehat{j}}{k + 1}$$

Also, $$|\vec b| \leq \sqrt{37}$$
$$\Rightarrow \displaystyle \dfrac{1}{k+1} \sqrt{25 + 25 k^2} \leq \sqrt{37}$$
or $$5\sqrt{1 + k^2} \leq \sqrt{37} (k + 1)$$

Squaring both sides, we get
$$25 (1 + k^2) \leq 37 (k^2 + 2k +1)$$

or $$6k^2 + 37k + 6 \geq 0$$ or $$(6k + 1) (k + 6) \geq 0$$

$$k \in (-\infty, - 6] \cup \left [ - \dfrac{1}{6}, \infty \right )$$
Question 3
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Vectors $$\vec a = \hat i + 2 \hat j + 3 \hat k, \vec b = 2 \hat i - \hat j + \hat k$$ and $$\vec c = 3 \hat i + \hat j + 4 \hat k$$ are so placed that the end point of one vector is the starting point of the next vector, then the vectors are

A
Not coplanar

B
Coplanar but cannot form a triangle

C
Coplanar and form a triangle

D
Coplanar and can form a right-angled triangle

Solution

Given $$\vec{AB}=\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$
$$\vec{BC}=\vec{b}=2\hat{i}-\hat{j}+\hat{k}$$
$$\vec{AC}=\vec{c}=3\hat{i}+\hat{j}+4\hat{k}$$
To check coplanar $$[\vec{a}\vec{b}\vec{c}]=0$$
\begin{vmatrix}1&2&3\\ 2&-1&1\\ 3&1&4 \end{vmatrix}=0
$$1(-5)-2(5)+3(5)=0$$
It is coplanar
Now to check a triangular
$$\vec{AC}=\vec{AB}+\vec{BC}$$
$$\hat{i}+2\hat{j}+3\hat{k}+2\hat{i}-\hat{j}+\hat{k}=\vec{AC}$$
$$3\hat{i}+\hat{j}+4\hat{k}=3\hat{i}+\hat{j}+4\hat{k}=\vec{AC}$$
So condition is followed it is triangular
Question 4
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$$L_{1}and L_{2}$$ are two lines whose vector equations are

$$L_{1}:\vec{r}=\lambda \left ( (\cos \theta+\sqrt{3})\hat{i}+(\sqrt{2}\sin\theta)\hat{j}+(\cos \theta-\sqrt{3})\hat{k} \right )$$

$$L_{2}:\vec{r}=\mu \left ( a \hat{i}+b \hat{j}+c\hat{k} \right ),$$

Where $$\lambda\ and\ \mu $$ are scalars and $$\alpha$$ is the acute angle

between $$L_{1}\ and\ L_{2}$$ . If the angle '$$\alpha$$'is independent

of $$\theta $$ then the value of '$$\alpha$$ ' is

A
$$\dfrac{\pi}{6}$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{3}$$

D
$$\dfrac{\pi}{2}$$

Solution

we know that $$cos\alpha=\dfrac{L_1.L_2}{|L_1|.|L_2|}$$ where $$\alpha$$ is the angle between the vectors $$L_1,L_2$$.

$$\implies cos\alpha=\dfrac{\lambda\mu(a(cos\theta+\sqrt 3)+b(\sqrt 2sin\theta)+c(cos\theta-\sqrt 3))}{\lambda\mu(\sqrt{a^2+b^2+c^2}).(\sqrt{cos^2\theta+3+2\sqrt 3cos\theta+2sin^2\theta+cos^2\theta+3-2\sqrt 3cos\theta})}$$

$$\implies cos\alpha=\dfrac{(a+c)cos\theta+\sqrt 3(a-c)+b\sqrt 2sin\theta}{8(\sqrt{a^2+b^2+c^2})}$$

For $$\alpha$$ to be independent of $$\theta$$

$$a+c=0$$ ..........(1)

and $$b=0$$ ........(2)

Therefore, $$cos\alpha=\dfrac{\sqrt 3(a-c)}{\sqrt 8.(\sqrt{a^2+c^2})}$$

$$\implies cos\alpha=\dfrac{\sqrt 3(2a)}{(\sqrt{8\times2a^2})}$$

$$\implies cos\alpha=\dfrac{\sqrt 3}{2}$$

$$\implies \alpha=\dfrac{\pi}{6}$$
Question 5
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Directions For Questions

$$ABC$$ is a triangle in which $$\overrightarrow{AB}=\vec {b}$$ and $$ \overrightarrow{AC}=\vec {c}.P$$ is a point on $$AB$$ such that $$AP : PB = 1 : 2$$.
$$Q$$ is a point on $$BC$$ such that $$CQ : QB = 2 : 1$$. $$AQ$$ and $$CP$$ meet at $$R$$.

...view full instructions

$$\overrightarrow{AR}$$ is

A
$$\dfrac{1}{5}(2\vec {b}+\vec {c})$$

B
$$\dfrac{1}{6}(2\vec {b}+\vec {c})$$

C
$$\dfrac{1}{7}(\vec {b}+2\vec {c})$$

D
$$\dfrac{1}{7}(2\vec {b}+\vec {c})$$

Solution

Choosing $$A$$ as the origin of reference , we find $$\bar{b}$$ and $$\, \bar{c}$$ are the position vectors of $$B$$ and $$C$$ respectively.

Position vector of $$P$$ is $$\displaystyle \dfrac{\bar{b}}{3}$$

Position vector of $$Q$$ is $$\displaystyle \dfrac{2\bar{b}+\bar{c}}{3}$$

$$\therefore $$ Lets take $$R$$ divides $$AQ$$ in the ratio $$(1-t):t
$$
$$\displaystyle \overline{r}=(1-t)\overline{o}+t\left(\dfrac{2\overline{b}+\overline{c}}{3}\right)$$ where $$t$$ is a scalar
$$\overrightarrow { r } =t\left( \dfrac { 2b+c }{ 3 } \right) $$

Lets take $$R$$ divides $$CP$$ in the ratio $$s:1-s $$ is
$$\overline{r}=(1-s)\overline{c}+s \overline{\dfrac{b}{3}}$$, where $$s$$ is a scalar

If $$AQ$$ and $$CP$$ are to intersect
$$\displaystyle t\left(\dfrac{2\overline{b}+\overline{c}}{3}\right)=(1-s)\overline{c}+s\overline{ \dfrac{b}{3}}$$ for same $$t$$ and $$s$$

Equating like components
$$\displaystyle \dfrac{2t}{3}=\dfrac{s}{3}$$ and $$\displaystyle \dfrac{t}{3}=1-s$$

On solving we get,
$$\displaystyle t=\dfrac{3}{7}$$ and $$\displaystyle s=\dfrac{6}{7}$$

$$\therefore $$ $$\displaystyle \overline{AR}=\dfrac{3}{7}\left(\dfrac{2\overline{b}+\overline{c}}{3}\right)=\dfrac{1}{7}(2\overline{b}+\overline{c})$$
Question 6
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$$ABCD$$ a parallelogram, $$A_1$$ and $$B_1$$ are the midpoints of sides $$BC$$ and $$CD$$, respectively. If $$\vec{AA_1} + \vec{AB_1} = \lambda \vec{AC}$$, then $$\lambda$$ is equal to

A
$$\displaystyle \dfrac{1}{2}$$

B
$$1$$

C
$$\displaystyle \dfrac{3}{2}$$

D
$$2$$

Solution

Let P.V. of $$A, B$$ and $$D$$ be $$\vec 0, \vec b,$$ and $$\vec d$$, respectively.
Then P.V. of $$C, \vec c = \vec b + \vec d$$
Also, P.V of $$A_1 = \vec b + \displaystyle \dfrac{\vec d}{2}$$
and P.V. of $$B_1 = \vec d + \displaystyle \dfrac{\vec b}{2}$$
$$\Rightarrow \vec{AA_1} + \vec{AB_1} = \displaystyle \dfrac{3}{2} (\vec b + \vec d) = \dfrac{3}{2} \vec{AC}$$
Question 7
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In a parallelogram $$OABC,$$ vectors $$\vec{a}, \vec{b}, \vec{c}$$ are, respectively, the position vectors of vertices $$A, B, C$$ with reference to $$O$$ as origin. A point $$E$$ is taken on the side $$BC$$ which divides it in the ratio of $$2 : 1$$. Also, the line segment $$AE$$ intersects the line bisecting the angle $$\angle$$AOC internally at point $$P$$. If $$CP$$ when extended meets $$AB$$ in point $$F,$$ then the position vector of point $$P$$ is

A
$$\displaystyle \dfrac{|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$

B
$$\displaystyle \dfrac{3|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$

C
$$\displaystyle \dfrac{2|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$

D
None of these

Solution

Let the position vector of $$A$$ and $$C$$ be $$\vec{a}$$ and $$\vec{c}$$ respectively. Therefore,
Position vector of $$B = \vec{b} = \vec{a} + \vec{c}$$...........................................................$$(i)$$
Also Position vector of $$E = \displaystyle \dfrac{\vec{b} + 2 \vec{c}}{3} = \dfrac{\vec{a} + 3 \vec{c}}{3}$$ ............................................................$$(ii)$$
Now point $$P$$ lies on angle bisector of $$\angle AOC.$$ Thus,
Position vector of point $$P =\displaystyle \lambda \left ( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right )$$............................................................................$$(iii)$$
Also let $$P$$ divides $$EA$$ in ration $$\mu : 1$$. Therefore,
Position vector of $$P$$
$$\displaystyle = \dfrac{\mu \vec{a} + \dfrac{\vec{a} + 3 \vec{c}}{3}}{\mu + 1} = \dfrac{(3 \mu + 1) \vec{a} + 3 \vec{c}}{3 (\mu + 1)}$$................................................................................$$(iv)$$
Comparing $$(iii)$$ and $$(iv),$$ we get
$$\lambda \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right ) = \dfrac{ (3 \mu + 1) \vec{a} + 3 \vec{c}}{3 (\mu + 1)}$$
$$\Rightarrow \displaystyle \dfrac{\lambda}{|\vec{a}|} = \dfrac{3 \mu + 1}{3 (\mu + 1)}$$ and $$\displaystyle \dfrac{\lambda}{|\vec{c}|} = \dfrac{1}{\mu + 1}$$
$$\Rightarrow \displaystyle \dfrac{3 |\vec{c}| - |\vec{a}|}{3 |\vec{a}|} = \mu$$
$$\displaystyle \Rightarrow \dfrac{\lambda}{|\vec{c}|} = \dfrac{1}{\dfrac{3 |\vec{c}| - \vec{a}}{3 |\vec{a}|}+1}$$
$$\Rightarrow \displaystyle \lambda = \dfrac{3 |\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|}$$
Hence, position vector of $$P$$ is $$\displaystyle \dfrac{3 |\vec{a}| |\vec{c}|}{ 3 |\vec{c}| + 2 |\vec{c}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$
Let $$F$$ divides $$AB$$ in ratio $$t : 1,$$ then position vector of $$F$$ is $$\displaystyle \dfrac{t \vec{b} + \vec{a}}{ t + 1} $$
Question 8
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Directions For Questions

Let $$OABCD$$ be a pentagon in which the sides $$OA$$ and $$CB$$ are parallel and the sides $$OD$$ and $$AB$$ are parallel. Also $$OA : CB =2 : 1$$ and $$OD : AB = 1 : 3$$.

...view full instructions

The ratio $$\displaystyle \dfrac{OX}{XC}$$ is

A
$$\dfrac{3}{4}$$

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

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

D
$$\dfrac{1}{2}$$

Solution

Let the position vectors of $$A, B, C$$ and $$D$$ be $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$, respectively.
Then, $$OA : CB = 2 : 1$$
$$\Rightarrow \vec{OA} = 2 \vec{CB}$$
$$\Rightarrow \vec{a} = 2 (\vec{b} - \vec{c})$$ ....$$(i)$$
and $$OD : AB = 1 : 3$$
$$3\vec{OD} = \vec{AB}$$
$$\Rightarrow 3\vec{d} = (\vec{b} - \vec{a}) = \vec{b} - 2 (\vec{b} - \vec{c})$$ [Using $$(i)$$]
$$= -\vec{b}+ 2\vec{c}$$ .....$$(ii)$$

Let $$OX : XC = \lambda : 1$$ and $$AX : XD =\mu : 1$$
Now,
$$X$$ divides $$OC$$ in the ratio $$\lambda : 1$$.
Therefore,
P.V of $$X = \displaystyle \frac{\lambda \vec{c}}{\lambda + 1}$$ ....$$(iii)$$

$$X$$ also divides $$AD$$ in the ratio $$\mu : 1$$.
Therefore,
P.V. of $$\displaystyle X = \dfrac{\mu \vec{d} + \vec{a}}{\mu + 1}$$ ...$$(iv)$$

From $$(iii)$$ and $$(iv)$$, we get
$$\displaystyle \dfrac{\lambda \vec{c}}{\lambda + 1} = \dfrac{\mu \vec{d} + \vec{a}}{\mu + 1}$$

or $$\displaystyle \left ( \frac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \frac{\mu}{\mu + 1} \right )\vec{d} + \left ( \frac{1}{\mu + 1} \right ) \vec{a}$$

or $$\displaystyle \left (\dfrac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \dfrac{\mu}{\mu + 1} \right ) \left ( \dfrac{-\vec{b} + 2 \vec{c}}{3} \right ) + \left ( \dfrac{1}{\mu + 1} \right ) 2 \left ( \vec{b} - \vec{c} \right )$$ ....(using $$(i)$$ and $$(ii)$$)

or $$\displaystyle \left ( \frac{\lambda}{\lambda + 1} \right ) \vec{c}= \left ( \frac{6 - \mu}{3 (\mu + 1)} \right ) \vec{b} + \left ( \frac{2 \mu}{3 (\mu + 1)} - \frac{2}{\mu + 1} \right ) \vec{c}$$

or $$\displaystyle \left ( \dfrac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \dfrac{6 - \mu}{3 (\mu + 1)} \right ) \vec{b} + \left ( \dfrac{2\mu - 6}{3 (\mu + 1)} \right ) \vec{c}$$

or $$\displaystyle \left ( \frac{6 - \mu}{3 (\mu + 1)} \right )\vec{b} + \left ( \frac{2 \mu - 6}{3 (\mu + 1)} - \frac{\lambda}{\lambda + 1} \right ) \vec{c} = \vec{0}$$

or $$\displaystyle \dfrac{6 - \mu}{3 (\mu + 1)} = 0$$ and $$\displaystyle \dfrac{2 \mu - 6}{3 (\mu + 1)} - \dfrac{\lambda}{\lambda + 1} = 0$$

(as $$\vec{b}$$ and $$\vec{c}$$ are non-collinear)

or $$\displaystyle \mu = 6, \lambda = \dfrac{2}{5}$$

Hence, $$ OX : XC = 2 : 5$$
Question 9
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The projection of the line joining the points $$(3, 4, 5)$$ and $$(4, 6, 3)$$ on the line joining the points $$(-1, 2, 4)$$ and $$(1, 0, 5)$$ is

A
$$\dfrac{4}{3}$$

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

C
$$\dfrac{1}{3}$$

D
$$\dfrac{1}{2}$$

Solution

Let $$AB=(4-3)\hat i+(6-4)\hat j+(3-5)\hat k=\hat i+2\hat j-2\hat k$$ and $$PQ=(1+1)\hat i+(0-2)\hat j+(5-4)\hat k=2\hat i-2\hat j+\hat k$$
Hence projection of $$AB$$ on $$PQ$$ is $$=\left |\cfrac{AB\cdot PQ}{|AB||PQ|}\right |$$
$$=\left |\cfrac{1(2)+2(-2)-2(1)}{\sqrt{3}\sqrt{3}}\right |$$
$$=\cfrac{4}{3}$$
Question 10
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A parallelogram is constructed on the vectors $$\bar{\alpha }$$ and $$\bar{\beta }$$. A vector which coincides with the altitude of the parallelogram and perpendicular to the side $$\bar{\alpha }$$ expressed in terms of the vectors $$\bar{\alpha }$$ and $$\bar{\beta }$$ is

A
$$\displaystyle \bar{\beta }+\dfrac{\bar{\beta }-\bar{\alpha }}{\left ( \bar{\alpha } \right )^{2}}\bar{\alpha }$$

B
$$\displaystyle \dfrac{\left ( \bar{\alpha }\times \bar{\beta } \right )\times \bar{\alpha }}{\left ( \bar{\alpha } \right )^{2}}$$

C
$$\displaystyle \dfrac{\bar{\beta }\cdot \bar{\alpha }}{\left ( \alpha \right )^{2}}\bar{\alpha }+\bar{\beta }$$

D
$$\displaystyle \left | \bar{\beta } \right |\dfrac{\bar{\alpha }\times \left ( \bar{\alpha }\times \bar{\beta } \right )}{\left ( \alpha \right )^{2}}$$

Solution

$$(\alpha\times\beta)$$ will give us the perpendicular vector or the normal to the plane where the parallelogram is lying.

Now
$$(\alpha\times \beta)\times \alpha$$ will give the vector parallel to the altitude of the parallelogram which is perpendicular to side $$\bar {\alpha}$$.

Thus the unit vector of the altitude of the given parallelogram perpendicular to $$\bar{\alpha}$$ will be
$$=\dfrac{(\alpha\times\beta)\times\alpha}{|\alpha|^{2}}$$