Vector Algebra Test

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Vector Algebra Test - 63

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Question 1
1 / -0

If $$\vec { x } $$ is a vector in the direction of $$(2,-2,1)$$ of magnitude $$6$$ and $$\vec { y } $$ is a vector in the direction of $$(1,1,-1)$$ of magnitude $$\sqrt{3}$$, then $$\left| \vec { x } +2\vec { y } \right| =...$$

A
$$40$$

B
$$\sqrt { 35 } $$

C
$$\sqrt { 17 } $$

D
$$2\sqrt { 10 } $$

Solution

they given x direction
we need to find unit vector in that direction and multiply with the magnitude of x
they given y direction
we need to find unit vector in that direction and multiply with the magnitude of y
$$\vec { x } =\cfrac { 6\left( 2\hat { i } -2\hat { j } +\hat { k } \right) }{ 3 }$$
$$ ,\vec { y } =\cfrac { \sqrt { 3 } \left( \hat { i } +\hat { j } -\hat { k } \right) }{ \sqrt { 3 } } $$
so $$\left| \vec { x } +2\vec { y } \right| $$
$$=\left| 6\hat { i } -2\hat { j } \right| =\sqrt { 40 } $$
$$=2\sqrt { 10 } $$
Question 2
1 / -0

The position vector of a point P is $$\overrightarrow r=x \overrightarrow i + y \overrightarrow j+x \overrightarrow k$$, Where $$x,y,z,\epsilon N$$ and $$\overrightarrow a= \overrightarrow i+ \overrightarrow j+\overrightarrow k$$. If $$\overrightarrow r. \overrightarrow a=10$$, then the number of possible positions of P is ___________.

A
$$30$$

B
$$72$$

C
$$66$$

D
$$36$$

Solution

The number of positive integral solution of the equation $$x_1+x_2+x_3+...x_r=n$$ is $$^{n-1}C_{r-1}$$
We have $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$
Also,$$\vec{r}.\vec{a}=10$$
$$\Rightarrow\,x+y+z=10$$
Thus number of positive integral solution is $$= ^{10-1}C_{3-1}= ^{9}C_{2}=\dfrac{9!}{7!2!}=\dfrac{9\times 8}{2}=36$$
Option$$(d)$$ is correct.
Question 3
1 / -0

If the position vector $$\vec{a}$$ of point $$(12, n) $$ is such that $$\left | \vec{a} \right | = 13$$, then find the value (s) of $$n$$.

A
$$\pm 6$$

B
$$\pm 4$$

C
$$\pm 5$$

D
$$\pm 7$$

Solution
Question 4
1 / -0

If a unit vector $$ \vec{a} $$ makes an angle $$ \dfrac{\pi }{3} $$ with $$ \hat{i},\dfrac{\pi }{4} $$ with $$ \hat{j} $$ and an accute angle $$ \theta $$ with $$ \hat{k}, $$ then find $$ \theta $$ and hence, the components of $$ \vec{a} $$.

A
$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$

B
$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{-1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$

C
$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$

D
$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}-\dfrac{1}{2}\hat{k} $$

Solution

The direction cosines of vector $$ \vec{a} $$ are $$ l = cos \dfrac{\pi }{3} = \dfrac{1}{2},m = cos \dfrac{\pi }{4} = \dfrac{1}{\sqrt{2}} $$ and $$ n = cos \,\theta $$.
$$ \therefore l^{2}+m^{2}+n = 1 \Rightarrow \dfrac{1}{4}+\dfrac{1}{2}+n^{2} = 1 \Rightarrow n^{2} = \dfrac{1}{4} \Rightarrow n = \dfrac{1}{2}\Rightarrow cos\,\theta = \dfrac{1}{2}\Rightarrow \theta = \dfrac{\pi }{3} $$
Since $$ \vec{a} $$ is a unit vector.
$$ \therefore \vec{a} = l\hat{i}+m\hat{j}+n\hat{k} $$
$$ \Rightarrow \vec{a} = \dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$
Hence, components of $$ \vec{a} $$ are $$ \dfrac{1}{2}\hat{i},\dfrac{1}{\sqrt{2}}\hat{j},\dfrac{1}{2}\hat{k}. $$
Question 5
1 / -0

What is the scalar projection of
$$\vec{a}=\hat{i}+2\hat{j}+\hat{k}$$ on $$\vec{b}=4\hat{i}+4\hat{j}+7\hat{k}
$$ ?

A
$$\dfrac{\sqrt{6}}{9}$$

B
$$\dfrac{19}{9}$$

C
$$\dfrac{9}{19}$$

D
$$\dfrac{\sqrt{6}}{19}$$

Solution

Formula,

scalar projection of a on b $$=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$$

Given,

$$a=i+2j+k,b=4i+4j+7k$$

$$=\dfrac{(i+2i+k)\cdot (4i+4j+7k)}{|4i-4j+7k|}$$

$$=\dfrac{4+8+7}{\sqrt{4^2+(4)^2+7^2}}$$

$$=\dfrac{19}{9}$$
Question 6
1 / -0

The adjacent sides of a parallelogram are represented by the vectors $$ \vec{a} = \hat{i}+\hat{j}+\hat{k} $$ and $$ \vec{b} = 2\hat{i}+\hat{j}+2\hat{k}.$$ Find unit vectors parallel to the diagonals of the parallelogram.

A
$$ \dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}+\hat{k}) $$

B
$$ \dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k}) $$

C
$$ \dfrac{1}{\sqrt{22}}(3\hat{i}+2\hat{j}+3\hat{k}),\dfrac{1}{\sqrt{2}}(\hat{i}+\hat{k}) $$

D
$$ \dfrac{1}{\sqrt{2}}(+\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k}) $$

Solution

Gven $$\vec a=\hat i+\hat j+\hat k$$ and $$\vec b=2\hat i+\hat j+2\hat k$$

From the figure

$$\vec c=\vec a+\vec b=\hat i+\hat j+\hat k+2\hat i+\hat j+2\hat k=3\hat i+2\hat j+3\hat k$$

and $$\vec a+\vec d=\vec b$$

$$\Rightarrow\vec d=\vec b-\vec a=2\hat i+\hat j+2\hat k-(\hat i+\hat j+\hat k)=\hat i+\hat k$$

The unit vector parallel to $$\vec c$$ is $$\dfrac{\vec c}{|\vec c|}=\dfrac{3\vec i+2\vec j+3\vec k}{\sqrt{3^2+2^2+3^2}}=\dfrac{1}{\sqrt{22}}(3\vec i+2\vec j+3\vec k)$$

The unit vector parallel to $$\vec d$$ is $$\dfrac{\vec d}{|\vec d|}=\dfrac{\vec i+\vec k}{\sqrt{(1)^2+(1)^2}}=\dfrac{1}{\sqrt{2}}(\vec i+\vec k)$$
Question 7
1 / -0

Express $$ \vec{AB}$$ in terms of unit vectors $$ \hat{i} $$ and $$\hat{j}$$, when the points are:
A(4,-1), B(1,3)
Find $$ \left | \vec{AB} \right |$$ in each case.

A
$$ \vec{AB} = -3\hat{i}-4\hat{j}, \left | \vec{AB} \right | = 5 $$

B
$$ \vec{AB} = +3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5 $$

C
$$ \vec{AB} = -3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5 $$

D
none of these

Solution

$$A(4, -1)$$ $$B(1, 3)$$
$$\vec{OA}=4\hat{i}-\hat{j}$$; $$\vec{OB}=\hat{i}+3\hat{j}$$
$$\vec{AD}=\vec{OB}-\vec{OA}=(\hat{i}+3\hat{j})-(4\hat{i}-\hat{j})=-3\hat{i}+4\hat{j}$$
$$|\vec{AB}|=\sqrt{(-3)^2+4^2}=5$$.
Question 8
1 / -0

The unit vector normal to the plane containing $$\vec{a}=(\hat{i}-\hat{j}-\hat{k})$$ and $$\vec{b}=(\hat{i}+\hat{j}+\hat{k})$$ is?

A
$$(\hat{j}-\hat{k})$$

B
$$(-\hat{j}+\hat{k})$$

C
$$\dfrac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$$

D
$$\dfrac{1}{\sqrt{2}}(-\hat{i}+\hat{k})$$

Solution
Question 9
1 / -0

If $$\bar{a}$$ and $$\bar{b} = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$$ are collinear and $$\bar{a} . \bar{b} = 27$$, then $$\bar{a}$$ is equal to

A
$$3 (\hat{i} + \hat{j} + \hat{k})$$

B
$$\hat{i} + 2\hat{j} + 2 \hat{k}$$

C
$$2 \hat{i} + 2\hat{j} + 2 \hat{k}$$

D
$$\hat{i} + 3\hat{j} + 3 \hat{k}$$

E
$$\hat{i} - 3 \hat{j} + 2 \hat{k}$$

Solution

Since, $$\vec { a }$$ and $$\vec { b }$$ are collinear vector. Therefore,
$$\vec { a }=\lambda\vec { b }$$............(i)
$$\because \vec { a }.\vec { b }=27$$
$$\Rightarrow |\vec { a }||\vec { b }|\cos 0^0=27$$
$$\Rightarrow |\vec { b }|.\sqrt{9+36+39}=27$$
$$\Rightarrow |\vec {a }|=\dfrac{27}{9}=3$$
By Equation (i),
$$\vec { }=\lambda \vec { b }$$
$$\Rightarrow |\vec { a}|=|\lambda||\vec { b }|$$
$$3=|\lambda|.9$$
$$\Rightarrow |\lambda|=\pm\dfrac{1}{3}$$
$$\therefore \vec { a }=\pm \dfrac{1}{3}(3\hat{i}+6\hat{j}+6\hat{k})$$
$$\vec { }=\pm (\hat{i}+2\hat{j}+2\hat{k})$$
Question 10
1 / -0

Let $$O$$ be the circumcentre, $$G$$ be the centroid and $$O$$ be the orthocentre of a $$\triangle ABC$$. Three vectors are taken through $$O$$ and are represented by $$\vec{a}=\vec{OA}, \vec{b}=\vec{OB}$$ and $$\vec{c}=\vec{OC}$$ then $$\vec{a}+\vec{b}+\vec{c}$$ is

A
$$\vec{OG}$$

B
$$2\vec{OG}$$

C
$$\vec{OO}$$

D
None of them

Solution

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 63

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Vector Algebra Test - 63

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SHARING IS CARING

Question 1

$$40$$

$$\sqrt { 35 } $$

$$\sqrt { 17 } $$

$$2\sqrt { 10 } $$

Solution

Question 2

$$30$$

$$72$$

$$66$$

$$36$$

Solution

Question 3

$$\pm 6$$

$$\pm 4$$

$$\pm 5$$

$$\pm 7$$

Solution

Question 4

$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$

$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{-1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$

$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k} $$

$$ \dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}-\dfrac{1}{2}\hat{k} $$

Solution

Question 5

$$\dfrac{\sqrt{6}}{9}$$

$$\dfrac{19}{9}$$

$$\dfrac{9}{19}$$

$$\dfrac{\sqrt{6}}{19}$$

Solution

Question 6

$$ \dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}+\hat{k}) $$

$$ \dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k}) $$

$$ \dfrac{1}{\sqrt{22}}(3\hat{i}+2\hat{j}+3\hat{k}),\dfrac{1}{\sqrt{2}}(\hat{i}+\hat{k}) $$

$$ \dfrac{1}{\sqrt{2}}(+\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k}) $$

Solution

Question 7

$$ \vec{AB} = -3\hat{i}-4\hat{j}, \left | \vec{AB} \right | = 5 $$

$$ \vec{AB} = +3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5 $$

$$ \vec{AB} = -3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5 $$

none of these

Solution

Question 8

$$(\hat{j}-\hat{k})$$

$$(-\hat{j}+\hat{k})$$

$$\dfrac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$$

$$\dfrac{1}{\sqrt{2}}(-\hat{i}+\hat{k})$$

Solution

Question 9

$$3 (\hat{i} + \hat{j} + \hat{k})$$

$$\hat{i} + 2\hat{j} + 2 \hat{k}$$

$$2 \hat{i} + 2\hat{j} + 2 \hat{k}$$

$$\hat{i} + 3\hat{j} + 3 \hat{k}$$

$$\hat{i} - 3 \hat{j} + 2 \hat{k}$$

Solution

Question 10

$$\vec{OG}$$

$$2\vec{OG}$$

$$\vec{OO}$$

None of them

Solution

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