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

We have, a = 2 î + ĵ 2 k̂ ⇒ |a| = √(4+1+4)=3 and b=î+ ĵ⇒ |b| = √(1+1)=√(2) Now, |c a|=3 ⇒ |c a|^2=9 ⇒ (c a).(c a)=9 ⇒ |c|^2+|a|^2 2c.a=9 Again, |(a × b) × c|= ...

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Byju's Answer

Standard XIII

Mathematics

Cross Product

Let a = 2 î+ ...

Question

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

$\frac{25}{8}$

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$\frac{1}{8}$

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Solution

The correct option is B
2

We have,
$a = 2^i +^j - 2^k \Rightarrow | a | = \sqrt{4 + 1 + 4} = 3 and b =^i +^j \Rightarrow | b | = \sqrt{1 + 1} = \sqrt{2} Now, | c - a | = 3 \Rightarrow | c - a |^{2} = 9 \Rightarrow (c - a) . (c - a) = 9 \Rightarrow | c |^{2} + | a |^{2} - 2 c . a = 9 Again, | (a \times b) \times c | = 3 \Rightarrow | a \times b | | c | s i n 30^{o} = 3 \Rightarrow | c | = \frac{6}{| a \times b |} But a \times b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & 1 & - 2 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2^i - 2^j +^k ∴ | c | = \frac{6}{\sqrt{4 + 4 + 1}} = 2$
FromEqs. (i) and (ii), we get
$\Rightarrow (2)^{2} + (3)^{2} - 2 c . a = 9 \Rightarrow c . a = 2$

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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 $a = 2^i +^j - 2^k, b =^i +^j$ 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