Vector Algebra Test

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Vector Algebra Test - 65

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Question 1
1 / -0

Let $$\hat{a} \space and \space \hat{b}$$ be mutually perpendicular unit vectors. Then for any arbitrary $$\vec{r}$$.

A
$$\vec{r} = (\vec{r} . \hat{a})\hat{a} + (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$$

B
$$\vec{r} = (\vec{r} . \hat{a}) - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b}))) (\hat{a} \times \hat{b})$$

C
$$\vec{r} = (\vec{r} . \hat{a})\hat{a} - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$$

D
none of these
Question 2
1 / -0

The vector with initial point $$P(2,-3,5)$$ and terminal point $$Q(3,-4,7)$$ is

A
$$\hat i-\hat j+2\hat k$$

B
$$5\hat i-7\hat j+12\hat k$$

C
$$\hat i+\hat j-2\hat k$$

D
$$None\ of\ these$$

Solution

$$(A)$$ is the correct answer.
Required vector $$=|\vec{AB}|=(3-2) \hat i+(-4+3)\hat j+(7-5)\hat k$$
$$=\hat i-\hat j+2\hat k$$
Question 3
1 / -0

The position vector of the point which divides the join of points with position vectors $$\vec a +\vec b$$ and $$2\vec a-\vec b$$ in the ratio $$1:2$$ is

A
$$\dfrac {3\vec a+2\vec b}{3}$$

B
$$\vec a$$

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

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

Solution

$$(D)$$ is the correct answer. Applying section formula, the position vector of the required point is $$\dfrac {2(\vec a+\vec b)+1(2\vec a-\vec b)}{2+1}=\dfrac {4\vec a+\vec b}{3}$$
Since, the position vector of a point $$R$$ divides the line segment joining the points $$P$$ and $$Q$$, whose position vectors are $$\vec p$$ and $$\vec q$$ in the ration $$m:n$$ internally, is given by $$\dfrac{m\vec q +n\vec p}{m+n}$$
Question 4
1 / -0

Let $$ \vec{u}, \vec{v} $$ and $$ \vec{w} $$ be vectors such that $$ \vec{u}+\vec{v}+\vec{w}=0 . $$ If $$ |\vec{u}|=3,|\vec{v}|=4 $$ and $$ |\vec{w}|=5, $$ then $$ \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u} $$ is

A
$$47$$

B
$$-25$$

C
$$0$$

D
$$25$$

E
$$50$$

Solution
Question 5
1 / -0

Line $$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$$ will not meet the plane $$\overrightarrow{r} \cdot \overrightarrow{n} = q$$, if

A
$$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$$

B
$$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$$

C
$$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$$

D
$$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$$

Solution

Given line is $$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$$
Substitute it in plane equation $$\overrightarrow{r} \cdot \overrightarrow{n} = q$$,
$$( \overrightarrow{a} + \lambda \overrightarrow{b}).\overrightarrow{n}=q$$
$$( \overrightarrow{a}. \overrightarrow{n}+ \lambda \overrightarrow{b}.\overrightarrow{n})=q$$

We must have $$\overrightarrow{b} \cdot \overrightarrow{n} = 0$$
and $$\overrightarrow{a} \cdot \overrightarrow{n} \neq q$$ (as point $$\overrightarrow{a} $$ on the line should not lie on the plane.
Question 6
1 / -0

If $$\vec \alpha | =4$$ and $$ -3 \le \lambda \le 2$$, then the range of $$ | \lambda \vec \alpha |$$ is

A
$$[0, 8]$$

B
$$[-12, 8]$$

C
$$[0, 12]$$

D
$$[8, 12]$$

Solution

We have, $$| \vec \alpha | =4$$ and $$-3 \le \lambda \le 2$$

$$\therefore | \lambda \vec a| =|-3| 4=12$$, at $$\lambda =-3$$

$$| \lambda \vec a| =|0|4=0$$, at $$\lambda =0$$

And $$| \lambda \vec \alpha | |2| 4=8$$, at $$\lambda =2$$

So, the range of $$| \lambda \vec \alpha |$$ is $$[0, 12]$$.

Alternate Method
Since, $$-3 \le \lambda \le 2$$

$$0 \le || \lambda | \le 3$$

$$\Rightarrow 0 \le 4 | \lambda | \le 12$$

$$| \lambda \vec a| \in [ 0, 12]$$
Question 7
1 / -0

If $$\vec a, \vec b, \vec c$$ are unit vector such that $$\vec a +\vec b +\vec c=\vec 0$$, then the value of $$\vec a \vec b+\vec b. \vec c+\vec c. \vec a$$ is

A
$$1$$

B
$$3$$

C
$$-\dfrac 32$$

D
$$None\ of\ these$$

Solution

We have, $$\vec a+ \vec b+ \vec c= \vec 0$$, and $$\vec a^2=1, \vec b^2 =1, \vec c^2 =1 | \vec a|=1, | \vec b|=1, | \vec c| =1$$

$$\because ( \vec a+ \vec b +\vec c)( \vec a+ \vec b +\vec c)=0$$

$$\Rightarrow \vec a^2 + \vec a. \vec b + \vec a. \vec c+ \vec +\vec b . \vec a + \vec b^2 + \vec b. \vec c+ \vec c. \vec a+ \vec c. \vec b +\vec c^2=0$$

$$\Rightarrow \vec a^2 +\vec b^2.+\vec c^2 +2( \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a)=0$$ $$[ \because \vec a. \vec b =\vec b. \vec a, \vec b. \vec c= \vec c. \vec b$$ and $$\vec c. \vec a= \vec a. \vec c]$$

$$\Rightarrow 1+1+1+2( \vec a. \vec b + \vec b. \vec c + \vec c. \vec a)=0$$

$$\Rightarrow \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a=-\dfrac 32$$
Question 8
1 / -0

The vector having initial and terminal points as $$(2, 5, 0)$$ and $$(-3, 7, 4)$$, respectively is

A
$$-\hat i +12 \hat j +4\hat k$$

B
$$5\hat i+2\hat j-4\hat k$$

C
$$-5\hat i +2\hat j+4\hat k$$

D
$$\hat i+\hat j+ \hat k$$

Solution

Required vector $$=|\vec{AB}|=(-3-2) \hat i+(7-5)\hat j+(4-0)\hat k$$
$$=-5\hat i+2\hat j=4\hat k$$
Question 9
1 / -0

The projection of vector $$\vec a=2\hat i-\hat j+\hat k$$ along $$\vec b=\hat i+2\hat j+2\hat k$$ is

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

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

C
$$2$$

D
$$\sqrt 6$$

Solution

$$(A)$$ is the correct answer. Projection of a vector $$\vec a$$ on $$\vec b$$ is
$$\dfrac {\vec a \vec b}{|b|}=\dfrac {(2\hat i-\hat j+\hat k)(\hat i+2\hat j+2\hat k)}{\sqrt {1+4+4}}=\dfrac 23$$
Question 10
1 / -0

If $$\vec a, \vec b, \vec c$$ are three vectors such that $$ \vec a +\vec b+ \vec c=\vec 0$$ and $$ | \vec a| =2, | \vec b|=3, | \vec c| =5$$, then value of $$\vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a$$ is

A
$$0$$

B
$$1$$

C
$$-19$$

D
$$38$$

Solution

Here, $$ \vec a+ \vec b+ \vec c=\vec 0$$ and $$\vec a^2 =4, \vec b^2 =9, \vec c^2=25$$

$$\therefore ( \vec a+ \vec b+ \vec c). ( \vec a+ \vec b+\vec c)=\vec 0$$

$$\Rightarrow \vec a^2 +\vec a. \vec b+ \vec a. \vec c+\vec b. \vec a+\vec b^2 + \vec b. \vec c+ \vec c. \vec a+ \vec c. \vec b+ \vec c^2=\vec 0$$

$$\Rightarrow \vec a^2+ \vec b^2+\vec c^2+2( \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a)=0$$ $$[ \because \vec a. \vec b=\vec b. \vec a]$$

$$\Rightarrow 4+9+25+2( \vec a. \vec b+ \vec b. \vec c +\vec c. \vec a)=0$$

$$\Rightarrow \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a=\dfrac{-38}{2}=-19$$

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Vector Algebra Test - 65

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Vector Algebra Test - 65

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SHARING IS CARING

Question 1

$$\vec{r} = (\vec{r} . \hat{a})\hat{a} + (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$$

$$\vec{r} = (\vec{r} . \hat{a}) - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b}))) (\hat{a} \times \hat{b})$$

$$\vec{r} = (\vec{r} . \hat{a})\hat{a} - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$$

none of these

Question 2

$$\hat i-\hat j+2\hat k$$

$$5\hat i-7\hat j+12\hat k$$

$$\hat i+\hat j-2\hat k$$

$$None\ of\ these$$

Solution

Question 3

$$\dfrac {3\vec a+2\vec b}{3}$$

$$\vec a$$

$$\dfrac {5\vec a-\vec b}{3}$$

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

Solution

Question 4

$$47$$

$$-25$$

$$0$$

$$25$$

$$50$$

Solution

Question 5

$$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$$

$$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$$

$$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$$

$$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$$

Solution

Question 6

$$[0, 8]$$

$$[-12, 8]$$

$$[0, 12]$$

$$[8, 12]$$

Solution

Question 7

$$1$$

$$3$$

$$-\dfrac 32$$

$$None\ of\ these$$

Solution

Question 8

$$-\hat i +12 \hat j +4\hat k$$

$$5\hat i+2\hat j-4\hat k$$

$$-5\hat i +2\hat j+4\hat k$$

$$\hat i+\hat j+ \hat k$$

Solution

Question 9

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

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

$$2$$

$$\sqrt 6$$

Solution

Question 10

$$0$$

$$1$$

$$-19$$

$$38$$

Solution

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