Let a-2i-j-2k and b-i-j- Let c be a vector such that -c-a-3- -a- b- c-3 and the angle between a- b and c is 30- Then a-c is equal to

Explanation for the correct answer:Finding the value of a.c :Given that a=2i+j 2k and b=i+j.So, Magnitude of a, |a|=√(2^2+2^2+1)=√(9) =3and Magnitude of b, |b| ...

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Standard XI

Mathematics

Applications of Cross Product

Let a=2i+j-2k...

Question

Let $\vec{a} = 2 \vec{i} + \vec{j} - 2 \vec{k}$ and $\vec{b} = \vec{i} + \vec{j}$ . Let $c$ be a vector such that $|c - a| = 3$ , $|(a \times b) \times c| = 3$ and the angle between $a \times b$ and $c$ is $30 °$ . Then $a . c$ is equal to

$2$

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$5$

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$\frac{1}{8}$

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$\frac{25}{8}$

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Solution

The correct option is A
$2$

Explanation for the correct answer:
Finding the value of $a . c$ :
Given that $\vec{a} = 2 \vec{i} + \vec{j} - 2 \vec{k}$ and $\vec{b} = \vec{i} + \vec{j}$ .
So, Magnitude of $\vec{a}$ , $|\vec{a}| = \sqrt{2^{2} + 2^{2} + 1} = \sqrt{9} = 3$
and Magnitude of $\vec{b}$ , $|\vec{b}| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
According to the cross product, $|\vec{m} \times \vec{n}| = |\vec{m}| |\vec{n}| \sin (\emptyset)$
So, $|(\vec{a} \times \vec{b}) \times \vec{c}| = |(\vec{a} \times \vec{b})| |\vec{c}| \sin (30) ° \to (i)$ [Since, the angle between $a \times b$ and $c$ is $30 °$ ]
But given, $|(\vec{a} \times \vec{b}) \times \vec{c}| = 3$ , so
$\vec{\Rightarrow a} \times \vec{b} = |\begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & - 2 \\ 1 & 1 & 0 \end{matrix}| = \vec{i} (0 + 2) - \vec{j} (0 + 2) + \vec{k} (2 - 1) = 2 \vec{i} - 2 \vec{j} + \vec{k}$
and
$|\vec{a} \times \vec{b}| = \sqrt{2^{2} + {(- 2)}^{2} + 1} = \sqrt{9} = 3$
Therefore, substituting all the values in equation $(i)$
$|(\vec{a} \times \vec{b}) \times \vec{c}| = |(\vec{a} \times \vec{b})| |\vec{c}| \sin (30) ° 3 = 3 \times |\vec{c}| \times \frac{1}{2} [∵ |(\vec{a} \times \vec{b})| = 3; |(\vec{a} \times \vec{b}) \times \vec{c}| = 3; \sin (30) ° = \frac{1}{2}] \Rightarrow |\vec{c}| = 2$
Given,
$|c - a| = 3 Takingsquareonbothsides \Rightarrow {(|\overset{⇀}{c} - \overset{⇀}{a}|)}^{2} = {(3)}^{2} \Rightarrow {|\overset{⇀}{c}|}^{2} + {|\overset{⇀}{a}|}^{2} - 2 \overset{⇀}{a} . \overset{⇀}{c} = 9 [∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}] \Rightarrow {(2)}^{2} + 9 - 2 \overset{⇀}{a} . \overset{⇀}{c} = 9 [∵ |\overset{⇀}{a}| = 3, |\overset{⇀}{c}| = 2] \Rightarrow 4 - 2 \overset{⇀}{a} . \overset{⇀}{c} = 0 \Rightarrow \overset{⇀}{a} . \overset{⇀}{c} = \frac{4}{2} \Rightarrow \overset{⇀}{a} . \overset{⇀}{c} = 2$
Therefore, the value of $a . c$ is $2$ .
Hence, option(A) is the correct answer.

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Let $a = 2^i +^j - 2^k, b =^i +^j$ and $c$ be a vector such that $| c - a | = 3, | (a \times b) \times c | = 3$ and the angle between $c$ and $a \times b$ is $30^{\circ}$ . Then, $a \cdot c$ is equal to

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Let a→=2i→+j→-2k→ and b→=i→+j→. Let c be a vector such that c-a=3, a×b×c=3 and the angle between a×b and c is 30°. Then a.c is equal to

Let $\vec{a} = 2 \vec{i} + \vec{j} - 2 \vec{k}$ and $\vec{b} = \vec{i} + \vec{j}$ . Let $c$ be a vector such that $|c - a| = 3$ , $|(a \times b) \times c| = 3$ and the angle between $a \times b$ and $c$ is $30 °$ . Then $a . c$ is equal to