Vector Algebra Test

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Vector Algebra Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

A unit vector $${\vec a}$$ in the plane of $$\vec = 2\hat i + \hat j$$ and $$\vec c = \hat i - \hat j + \hat k$$ is such that angle between $${\vec a}$$ and $${\vec b}$$ is the same angle between $${\vec a}$$ and $${\vec d}$$ where $$\vec d = \hat j + 2\hat k$$

A
$$\frac{{\hat i + \hat j + \hat k}}{{\sqrt 3 }}$$

B
$$\frac{{\hat i - \hat j + \hat k}}{{\sqrt 3 }}$$

C
$$\frac{{2\hat i + \hat j}}{{\sqrt 5 }}$$

D
$$\frac{{2\hat i - \hat j}}{{\sqrt 5 }}$$

Solution

Solution -
Let $$\vec{a}=\dfrac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}(x\hat{i}+y\hat{j}+z\hat{k})$$
$$\vec{a}.(\vec{b}\times \vec{c})=0$$
$$\vec{a}\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end{vmatrix}=\vec{a}.(\hat{i}-2\hat{j}-3\hat{k})$$
$$x-2y-3z=0$$. ____ (1)
Also
$$\dfrac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}=\dfrac{\vec{a}.\vec{d}}{\left | \vec{a} \right ||\vec{d}|}$$
$$\dfrac{2x+y}{\sqrt{5}}=\dfrac{y+2z}{\sqrt{5}}$$
$$x=z$$ ____ (2)
$$y=-x=-z$$ (from (1))
From options
$$\vec{a}=\dfrac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{1^{2}+1^{2}+1^{2}}}=\dfrac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$$
B is correct.
Question 2
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ABCD is a parallelogram. The position vectors of A and C are respectively, $$3\hat{i}+3\hat{j}+5\hat{k}$$ and $$\hat{i}-5\hat{j}-5\hat{k}$$. If M is the mid-point of the diagonal DB, then the magnitude of the projection of $$\vec{OM}$$ on $$\vec{OC}$$, where O is the origin is?

A
$$\dfrac{7}{\sqrt{50}}$$

B
$$7\sqrt{50}$$

C
$$\dfrac{7}{\sqrt{51}}$$

D
$$7\sqrt{51}$$

Solution

Given the position vectors of A and C are $$ 3\hat{i}+3\hat{j}+5\hat{k} $$
and $$ \hat{i}-5\hat{j}-5\hat{k} $$
In a parallelogram the midpoints of both the diagonals
are same. So, given M is the midpoints of DB
implies M is also the midpoint of AC.
So, $$ \vec{OM} = \dfrac{\vec{OA}+\vec{OC}}{2} $$
$$ = \dfrac{(3\hat{i}+3\hat{j}+5\hat{k})+(\hat{i}-5\hat{j}-5\hat{k})}{2} $$
$$ = 2\hat{i}-\hat{j} $$
The magnitude of projection $$ \vec{OM} $$ on $$ \vec{OC} $$
$$ = \dfrac{|\vec{OM}.\vec{OC}|}{|\vec{OC}|}=\dfrac{|2+5|}{\sqrt{1+25+25}} $$
$$ = \dfrac{7}{\sqrt{51}} $$
Question 3
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If the unit vectors $$\vec{e}_1 \, and \, \vec{e}_2$$ are inclined at an angle $$2 \theta \, and \, |\vec{e}_1 - \vec{e}_2| < 1$$, then for $$\theta \in [0, \pi] , \theta$$ may lie in the interval

A
$$\left[0, \dfrac{\pi}{6} \right]$$

B
$$\left[ \dfrac{\pi}{6} , \dfrac{\pi}{2} \right]$$

C
$$\left[ \dfrac{5 \pi}{6} , \pi \right]$$

D
$$\left[\dfrac{\pi}{2}, \dfrac{5 \pi}{6} \right]$$

Solution

$$|\hat a-\hat b|<1$$

$$|\hat a|^2+|\hat b|^2-2|\hat a||\hat b|\cos 2\theta <1$$

$$2(1-\cos 2\theta)<1$$

$$2\times 2\sin^2\theta <1$$

$$\sin^2\theta <\dfrac 14$$

$$\dfrac{-1}{2}< \sin \theta <\dfrac 12$$

$$\therefore \theta \in (0, \dfrac \pi6 )\cup (\dfrac{5\pi}{6}, \pi)$$
Question 4
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Let $$a, b$$ and $$c$$ be three unit vectors such that a is perpendicular to the plane of $$b$$ and $$c$$. if the angle between $$b$$ and $$c$$ is $$\dfrac { \pi }{ 3 }$$, then $$|a\times b-a\times c|^{2}$$ is equal to

A
$$1/3$$

B
$$1/2$$

C
$$1$$

D
$$2$$

Solution
Question 5
1 / -0

If $$[\vec{a}\vec{b}\vec{c}]=2$$, then find the value of $$[(\vec{a}+2\vec{b}-\vec{c}(\vec{a}-\vec{b})(\vec{a}-\vec{b}-\vec{c})]$$.

A
$$6$$

B
$$8$$

C
$$12$$

D
$$11$$

Solution

$$[(\vec{a}+2\vec{b}-\vec{c})(\vec{a}-\vec{b})(\vec{a}-\vec{b}-\vec{c})]$$

$$=\begin{vmatrix} 1 & 2 & -1\\ 1 & -1 & 0\\ 1 & -1 & -1\end{vmatrix}[\vec{a}\vec{b}\vec{c}]$$

$$=3[\vec{a}\vec{b}\vec{c}]=6$$
Question 6
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Let $$\vec{a}=-\hat{i}-\hat{k}, \vec{b}=-\hat{i}+\hat{j}$$ and $$\vec{c}=\hat{i}+2\hat{j}+3\hat{k}$$ be three given vectors. If $$\vec{r}$$ is a vector such that $$\vec{r}\times \vec{b}=\vec{c}\times \vec{b}$$ and $$\vec{r}\cdot\vec{a}=0$$, then the value of $$\vec{r}\cdot \vec{b}$$ is?

A
$$8$$

B
$$9$$

C
$$6$$

D
None of these

Solution

Given $$ \vec{a} = -\hat{i}-\hat{k},\vec{b} = -\hat{i}+\hat{j},\vec{c} = \hat{i}+2\hat{j}+3\hat{k} $$
$$ \vec{r}\times \vec{b} = \vec{c}\times \vec{b} $$ and $$ \vec{r}.\vec{a} = 0 $$
Cross product with $$ \vec{a} $$ on both sides.
$$ \Rightarrow (\vec{r}\times\vec{b})\times\vec{a} = (\vec{c}\times\vec{b})\vec{a} $$
$$ \Rightarrow (\vec{r}.\vec{a})\vec{b}-(\vec{b}.\vec{a})\vec{r}=(\vec{c}.\vec{a})\vec{b}-(\vec{b}.\vec{a})\vec{c} $$
$$ \vec{a}.\vec{b} = (-1)(-1)+(1)(0)+(-1)(0) = 1 $$
$$ \vec{b}.\vec{C} = (-1)(1)+(1)(2)+(0)(3) = 1 $$
$$ \vec{c}.\vec{a} = (-1)(1)+(0)(2)+(-1)(3) = -4 $$
$$ \Rightarrow -\vec{r} = -4\vec{b}-\vec{c} $$
$$ \Rightarrow \vec{r} = 4\vec{b}+\vec{c} $$
$$ = -4\hat{i}+4\hat{j}+\hat{i}+2\hat{j}+3\hat{k} $$
$$ = -3\hat{i}+6\hat{i}+3\hat{k} $$
$$ \therefore \vec{r}.\vec{b}(-3\hat{i}+6\hat{j}+3\hat{k}).(-\hat{i}+\hat{j}) $$
$$ = 3+6 $$
$$ = 9 $$
Question 7
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A non-zero vectors $$\overrightarrow{a}$$ is such that its projections along the vectors $$\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$$ and $$\dfrac{-\hat{i}+\hat{j}}{\sqrt{2}}$$ and $$\hat{k}$$ are equal then unit vector along $$\overrightarrow{a}$$ is

A
$$\dfrac{\sqrt{2}\hat{j}-\hat{k}}{\sqrt{3}}$$

B
$$\dfrac{\hat{j}-\sqrt{2}\hat{k}}{\sqrt{3}}$$

C
$$\dfrac{\sqrt{2}}{\sqrt{3}}\hat{j}+\dfrac{\hat{k}}{\sqrt{3}}$$

D
$$\dfrac{\hat{j}-\hat{k}}{\sqrt{2}}$$

Solution
Question 8
1 / -0

Let $$O$$ be the centre a regular hexagon $$ABCDEF$$ Then the magnitude of sum of the vectors $$\overline { OA }, \overline { OB} ,\overline { OC },\overline { OD },\overline { OE }, $$ and $$\overline { OF } $$ is

A
$$\sqrt3$$

B
$$0$$

C
$$2$$

D
$$1$$

Solution
Question 9
1 / -0

The length of the projection of the line segment joining points $$(5, -1, 4)$$ and $$(4, -1, 8)$$ on the plane $$x+y+z=7$$.

A
$$\dfrac{2}{\sqrt{3}}$$

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

C
$$\sqrt{5}$$

D
$$\sqrt{\dfrac{2}{3}}$$

Solution

Let $$A\left(5,-1,4\right)$$ and $$B\left(4,-1,3\right)$$ be a line
then $$\vec{AB}=\left(4-5\right)\hat{i}+\left(-1+1\right)\hat{j}+\left(3-4\right)\hat{k}$$
$$=-\hat{i}-\hat{j}$$
$$\left|AB\right|=\sqrt{{1}^{2}+{1}^{2}}=\sqrt{2}$$ units
Given:Equation of the plane is $$x+y+z=7$
Normal vector$$\vec{n}=\hat{i}+\hat{j}+\hat{k}$$
Equation of plane is $$\vec{r}.\hat{n}=d$$
$$\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(\hat{i}+\hat{j}+\hat{k}\right)=7$$
$$\left|n\right|=\sqrt{1+1+1}=\sqrt{3}$$
Length of Projection $$AB$$ along the plane is $$P=\left|AB\right|\cos{\left(90-\theta\right)}=\left|AB\right|\sin{\theta}$$
Here $$\cos{\theta}=\dfrac{\vec{AB}.\vec{n}}{\left|AB\right|.\left|\vec{n}\right|}$$
$$=\dfrac{\left(-\hat{i}-\hat{j}\right).\left(\hat{i}+\hat{j}+\hat{k}\right)}{\left(\sqrt{2}\right).\left(\sqrt{3}\right)}$$
$$=\dfrac{-1-1+0}{\sqrt{6}}=-\dfrac{2}{\sqrt{6}}\times\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2\sqrt{3}}=\dfrac{\sqrt{2}}{\sqrt{3}}$$
$$\Rightarrow\,\sin{\theta}=\sqrt{1-{\cos}^{2}{\theta}}=\sqrt{1-{\left(\dfrac{\sqrt{2}}{\sqrt{3}}\right)}^{2}}=\sqrt{1-\dfrac{2}{3}}=\dfrac{1}{\sqrt{3}}$$
Now,$$P=\left|AB\right|\sin{\theta}=\sqrt{2}\times\dfrac{1}{\sqrt{3}}=\sqrt{\dfrac{2}{3}}$$
$$\therefore\,$$Length of projection of line segment on the plane is $$\sqrt{\dfrac{2}{3}}$$
Question 10
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If $$a$$ and $$b$$ are unit vectors along $$OA, OB$$ and $$OC$$ bisects the angle $$AOB$$. The unit vector along $$OC$$ is

A
$$\dfrac{a+b}{2}$$

B
$$\dfrac{b-b}{2}$$

C
$$\dfrac{a+b}{|a+b|}$$

D
$$\dfrac{a-b}{|a-b|}$$

Solution

$$\overline{OC} = k\left(\dfrac{\overline{OA}+\overline{OB}}{2}\right)$$
$$\overline{OC}$$ unit vector $$=\dfrac{\dfrac{k}{2} (\overline{OA} + \overline{OB})}{\dfrac{k}{2} |\overline{OA} + \overline{OB}|}$$
$$=\dfrac{\overline{a} + \overline{b}}{|\overline{a }+\overline{b}| }$$