Vector Algebra Test

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Vector Algebra Test - 58

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Question 1
1 / -0

If $$\vec {a}$$ and $$\vec {b}$$ are in the plane which is perpendicular to the plane containing $$\vec {c}$$ and $$\vec {d}$$ then $$(\vec {a} \times \vec {b})\times (\vec {c} \times \vec {d})$$ is

A
$$0$$

B
$$1$$

C
$$\vec {a}.\vec {d}+\vec {b}.\vec {c}$$

D
$$none\ of\ these$$

Solution

$$\left( \vec { a } \times \vec { b } \right) $$ is a plane $$\bot $$ to both $$\vec { a } $$ and $$\vec { b } $$
$$\left( \vec { c } \times \vec { d } \right) $$ is a plane $$\bot$$ to both $$\vec { c } $$ and $$\vec { d } $$
Now $$\vec { a } $$ and $$\vec { b } $$ are in plane which is perpendicular to plane $$\vec { c } $$ and $$\vec { d } $$
$$\therefore$$ $$\vec { a } \times \vec { b } $$ is in with zero degree with $$\vec { c } \times \vec { d } $$
$$\therefore$$ $$\left( \vec { a } \times \vec { b } \right) \times \left( \vec { c } \times \vec { d } \right) =0$$
Question 2
1 / -0

If $$\overrightarrow { a } =2\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =3\hat { i } -4\hat { j } +2\hat { k } ,\overrightarrow { c } =\hat { i } -2\hat { j } +2\hat { k } $$ then the projection of $$\overrightarrow { a } +\overrightarrow { b } $$ on $$\overrightarrow { c } $$ is

A
$$\cfrac{17}{3}$$

B
$$\cfrac{5}{3}$$

C
$$\cfrac{4}{3}$$

D
$$\cfrac{17}{\sqrt{43}}$$

Solution

$$\vec{a}+\vec{b}=5\hat{i}-3\hat{j}+3\hat{k}$$
Projection of $$\vec{a}+\vec{b}$$ on $$\vec{c}$$ is $$\dfrac{(\vec{a}+\vec{b}).\vec{c}}{|\vec{c}|}$$
$$\dfrac{5+6+6}{\sqrt{1+4+4}}=\dfrac{17}{3}$$
Question 3
1 / -0

If $$A(6,3,2),B(5,1,4),C(3,-4,7), D(0,2,5)$$ are four points, then projection of $$CD$$ on $$AB$$ is

A
$$-\cfrac{13}{3}$$

B
$$-\cfrac{13}{7}$$

C
$$-\cfrac{3}{13}$$

D
$$-\cfrac{7}{13}$$
Question 4
1 / -0

If $$\vec a,\vec b$$ and $${\vec c}$$ are unit vectors, then $${\left| {\vec a + \vec b} \right|^2} + {\left| {\vec b - \vec c} \right|^2} + {\left| {\vec c - \vec a} \right|^2}$$ does NOT exceed

A
4

B
9

C
8

D
6

Solution

Given $$\bar {a}, \bar {b} $$ and $$\bar {c}$$ are unit vector
we need maximum value, so :-

Let $$\bar {a}= \hat {i} \bar {b}= \hat {j} $$ and $$ \bar {c} = \hat {k}$$
$$|\bar {a}+ \bar {b}|^{2}+ |\bar {b}-\bar {c}|^{2}+ |\bar {c}-\bar {a}|^{2}$$

$$|\hat{i}+ \hat{j}|^{2}+ |\hat{j}- \hat{k}|^{2}+ |\hat{k}- \hat{i}|^{2}$$
$$(\sqrt{1^{2}+1^{2}})^{2}+ (\sqrt{1^{2}+1^{2}})^{2}+ (\sqrt{1^{2}+1^{2}})^{2}$$

$$\Rightarrow 2+2+2=6$$
$$\therefore 6$$ is the maximum value.

$$\therefore |\bar {a}+ \bar {b}|+ |\bar {b} - \bar {c}| + |\bar {c}- \bar {a}|^{2}$$ does not exceed $$6$$
Question 5
1 / -0

If $$\left| {\vec a} \right| = 2,\left| {\vec b} \right| = 3$$ and $$\left| {2\vec a - \vec b} \right| = 5,$$ then $$\left| {2\vec a + \vec b} \right|$$ equals:

A
$$17$$

B
$$7$$

C
$$5$$

D
$$1$$

Solution

We have
$$\left| {\overrightarrow a } \right| = 2\,\,\,\,\,\left| {\overrightarrow b } \right| = 3\,\,\,$$
And, $$\left| {2\overrightarrow a - \overrightarrow b } \right| = 5$$
We have to find the value of $$\left| {2\overrightarrow a + \overrightarrow b } \right| $$
Now
$$\left| {2\overrightarrow a - \overrightarrow b } \right|.\left| {2\overrightarrow a - \overrightarrow b } \right| = 25$$
$$4{\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} - 4\overrightarrow a .\overrightarrow b = 25$$
$$25 - 4\overrightarrow a .\overrightarrow b = 25$$
$$\therefore \overrightarrow a .\overrightarrow b = 0$$
Then for $$\overrightarrow a \bot \overrightarrow b $$
$${\left| {2\overrightarrow a + \overrightarrow b } \right|^2} = \left( {2\overrightarrow a + \overrightarrow b } \right).\left( {2\overrightarrow a + \overrightarrow b } \right)$$
$$ = 4{\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + 4\overrightarrow a .\overrightarrow b $$
$$ = 4 \times 4 + 9 = 25$$

$$\therefore \left| {2\overrightarrow a + \overrightarrow b } \right| = 5$$
Hence, the option $$C$$ is the correct answer.
Question 6
1 / -0

If the vectors $$\overrightarrow { a } =\hat { i } -\hat { j } +2\hat { k } ;\overrightarrow { b } =2\hat { i } +4\hat { j } +\hat { k } ;\overrightarrow { c } =\lambda \hat { i } +\hat { j } +\mu \hat { k } $$ are mutually orthogonal, then $$(\lambda,\mu)=$$

A
$$(2,-3)$$

B
$$(-2,3)$$

C
$$(3,-2)$$

D
$$(-3,2)$$

Solution

Given $$ \vec{a} = \hat{i}-\hat{j}+2\hat{k}$$
$$ \vec{b} = 2\hat{i}+4\hat{j}+\hat{k}$$
$$ \vec{c} = \lambda \hat{i}+\hat{j}+\mu \hat{k}$$
since, vector's are mutually orthogonal
$$ \therefore \vec{a}.\vec{b} = 0, \vec{b}.\vec{c} = 0, \vec{c}.\vec{a} = 0 $$
$$ \vec{b}.\vec{c} = 0$$
$$ 2\lambda +4+\mu = 0$$ __ (1)
$$ \vec{c}.\vec{a} = 0$$
$$ \lambda -1 +2\mu = 0$$ __ (II)
$$ (1)\times 2$$
$$ 4\lambda +8+2\mu = 0$$
$$ \lambda -1+2\mu = 0$$
____________
$$ 3\lambda +9 = 0$$
$$ \Rightarrow \lambda = -3$$
$$ -3-1+2\mu = 0$$
$$ \Rightarrow \mu = 2 $$
Hence, $$ (\lambda ,\mu ) =(-3,2)$$
Question 7
1 / -0

If $$\left| \overrightarrow { a } \right| =3,\left| \overrightarrow { b } \right| =4$$, if $$\left( \overrightarrow { a } +\lambda \overrightarrow { b } \right) $$ is perpendicular to $$\left( \overrightarrow { a } -\lambda \overrightarrow { b } \right) $$ then $$\lambda =$$

A
$$\cfrac{9}{16}$$

B
$$\cfrac{3}{5}$$

C
$$\cfrac{3}{4}$$

D
$$\cfrac{4}{3}$$

Solution

Given $$(\vec {a}+\lambda\vec {b})\bot (\vec {a}-\lambda {\vec{b}})$$
$$\therefore \quad (\vec {a}+\lambda \vec {b})(\vec {a}-\lambda \vec {b})=0$$
$$\Rightarrow \quad |\vec {a}|^{2}-\lambda^{2}|\vec {b}|^{2}=0$$
$$\Rightarrow \quad 9-\lambda^{2}\cdot 16^{2}=0$$
$$\Rightarrow \quad \dfrac {9}{16}=\lambda^{2}$$
$$\therefore \quad \lambda=\pm \dfrac {3}{4}$$
$$\therefore \quad \lambda =\dfrac {3}{4},\lambda=-\dfrac {3}{4}$$
Question 8
1 / -0

The projection of the vector $$2\hat i + \hat j - 3\hat k$$ on the vector $$\hat i - 2\hat j - \hat k$$

A
$$ - \dfrac{3}{{\sqrt {14} }}$$

B
$$ \dfrac{3}{{\sqrt {14} }}$$

C
$$ - \sqrt {\dfrac{3}{2}} $$

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

Solution

Given that:
$$\vec{P}=2\hat i+\hat j -3\hat k$$
$$\vec Q=\hat i-2\hat j -\hat k$$
Now
We know that projection of $$\vec P$$ on $$\vec Q$$ is :$$\frac{\vec P . \vec Q}{|\vec Q|}$$
So,
$$\vec P.\vec Q=(2\hat i+\hat j-3\hat k).(\hat i-2\hat j-\hat k)$$
$$\vec P.\vec Q=2\times1+(1\times(-2))+(-3\times-1)$$
$$\vec P.\vec Q=2-2+3$$
$$\vec P.\vec Q=3$$
Now
$$|\vec Q|=\sqrt{1+4+1}=\sqrt6$$
So,
$$proj{\vec Q}{\vec P}=\dfrac{3}{\sqrt6}$$
$$proj{\vec Q}{\vec P}=\dfrac{\sqrt3}{\sqrt2}$$
Question 9
1 / -0

The position vectors of the points A,B,C are $$\overline { i } + 2 \overline { j } - \overline { k } , \overline { i } + \overline { j } + \overline { k } , 2 \overline { i } + 3 \overline { j } + 2 \overline { k }$$ respectively. If A is chosen as the origin then the position vectors of B and C are

A
$$\overline { i } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$$

B
$$\overline { j } + 2 \overline { k } , \vec { i } + \overline { j } + 3 \overline { k }$$

C
$$- \overline { j } + 2 \overline { k } , \vec { i } - \overline { j } + 3 \overline { k }$$

D
$$- \overline { j } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$$

Solution

$$\textbf{Step1-Find position vector from given point}$$
$$\text{The origin is shifted so the new position vectors of B and C would}$$
$$\text{ be the difference of the old position vectors and the new origin vector i.e.}$$
$$(i+j+k) - (i+2j-k) = -j+2k$$ ($$\text{ New position vector of point B}$$)
$$(2i+3j+2k) - (i+2j-k) = i+j+3k$$ ($$\text{ New position vector of point A}$$)
$$\textbf{Hence , option D is correct}$$
Question 10
1 / -0

If $$|a|=5.|\vec{b}|=4$$, and $$|c|=3$$. then what will be the value of $$\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}$$ given that $$\vec{a}+\vec{b}+\vec{c}=0$$

A
$$25$$

B
$$50$$

C
$$-25$$

D
$$-50$$

Solution

Given $$\left | \vec{a} \right |=5\left | \vec{b} \right |=4$$ $$\left | \vec{c} \right |=3$$ $$\vec{a}+\vec{b}+\vec{c} =0$$

$$\left | \vec{a}+\vec{b}+\vec{c} \right |=0$$

$$\Rightarrow \left | \vec{a}+\vec{b}+\vec{c} \right |^{2}=0=(\vec{a}+\vec{b}+\vec{c}).(\vec{a}+\vec{b}+\vec{c})$$

$$\Rightarrow \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{a}+\vec{b}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}+\vec{c}.\vec{b}+\vec{c}.\vec{c}=0$$

$$\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+\left | \vec{c} \right |^{2}+ 2(\vec{a}\vec{b}+\vec{b}\vec{c}+\vec{c}\vec{a})=0$$

$$25+16+9+2(\vec{a}\vec{b}+\vec{b}\vec{c}+\vec{c}\vec{a})=0$$

$$\Rightarrow \vec{a}\vec{b}+\vec{b}\vec{c}+\vec{c}\vec{a}=-25$$

$$\therefore $$ Option C is correct

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 58

Result

Vector Algebra Test - 58

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Rank

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$\vec {a}.\vec {d}+\vec {b}.\vec {c}$$

$$none\ of\ these$$

Solution

Question 2

$$\cfrac{17}{3}$$

$$\cfrac{5}{3}$$

$$\cfrac{4}{3}$$

$$\cfrac{17}{\sqrt{43}}$$

Solution

Question 3

$$-\cfrac{13}{3}$$

$$-\cfrac{13}{7}$$

$$-\cfrac{3}{13}$$

$$-\cfrac{7}{13}$$

Question 4

4

9

8

6

Solution

Question 5

$$17$$

$$7$$

$$5$$

$$1$$

Solution

Question 6

$$(2,-3)$$

$$(-2,3)$$

$$(3,-2)$$

$$(-3,2)$$

Solution

Question 7

$$\cfrac{9}{16}$$

$$\cfrac{3}{5}$$

$$\cfrac{3}{4}$$

$$\cfrac{4}{3}$$

Solution

Question 8

$$ - \dfrac{3}{{\sqrt {14} }}$$

$$ \dfrac{3}{{\sqrt {14} }}$$

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

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

Solution

Question 9

$$\overline { i } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$$

$$\overline { j } + 2 \overline { k } , \vec { i } + \overline { j } + 3 \overline { k }$$

$$- \overline { j } + 2 \overline { k } , \vec { i } - \overline { j } + 3 \overline { k }$$

$$- \overline { j } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$$

Solution

Question 10

$$25$$

$$50$$

$$-25$$

$$-50$$

Solution

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