Let a-2-2k- and b- If c is a vector such that a-c-c- -c-a-2-2 and the angle between a-b and c is -6- then the value of -a-b-c- is

A=2î+ĵ 2k̂ ⇒|a|=√(4+1+4)=3 Now, |c a|=2√(2) ⇒|c a|^2=8 ⇒|c|^2+|a|^2 2c·a=8 ⇒|c|^2+9 2|c|=8 (∵a·c=|c| ) ⇒ |c|^2 2|c| + 1 = 0 ⇒(|c| 1)^2 = 0 ∴|c|=1 And angle ...

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Byju's Answer

Standard XII

Physics

Summary

Let a=2î+ĵ-2k...

Question

Let $\to a = 2^i +^j - 2^k$ and $\to b =^i +^j .$ If $\to c$ is a vector such that $\to a \cdot \to c = ∣ ∣ \to c ∣ ∣,$ $∣ ∣ \to c - \to a ∣ ∣ = 2 \sqrt{2}$ and the angle between $(\to a \times \to b)$ and $\to c$ is $\frac{π}{6},$ then the value of $∣ ∣ ∣ (\to a \times \to b) \times \to c ∣ ∣ ∣$ is

4

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\frac{2}{3}

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\frac{3}{2}

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Solution

The correct option is C $\frac{3}{2}$
$\to a = 2^i +^j - 2^k$
$\Rightarrow ∣ ∣ \to a ∣ ∣ = \sqrt{4 + 1 + 4} = 3$
Now, $∣ ∣ \to c - \to a ∣ ∣ = 2 \sqrt{2}$
$\Rightarrow {∣ ∣ \to c - \to a ∣ ∣}^{2} = 8$
$\Rightarrow {∣ ∣ \to c ∣ ∣}^{2} + {∣ ∣ \to a ∣ ∣}^{2} - 2 \to c \cdot \to a = 8$
$\Rightarrow {∣ ∣ \to c ∣ ∣}^{2} + 9 - 2 ∣ ∣ \to c ∣ ∣ = 8 (∵ \to a \cdot \to c = ∣ ∣ \to c ∣ ∣)$
$\Rightarrow | \to c |^{2} - 2 | \to c | + 1 = 0$
$\Rightarrow {(∣ ∣ \to c ∣ ∣ - 1)}^{2} = 0$
$∴ ∣ ∣ \to c ∣ ∣ = 1$
And angle between $\to a \times \to b$ and $\to c$ is $θ = \frac{π}{6}$ then $∣ ∣ ∣ (\to a \times \to b) \times \to c ∣ ∣ ∣ = ∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣ \cdot ∣ ∣ \to c ∣ ∣ \cdot sin θ$
$= ∣ ∣ 2^i - 2^j +^k ∣ ∣ \cdot 1 \cdot sin \frac{π}{6}$
$= \frac{3}{2}$

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Let →a=2^i+^j−2^k and →b=^i+^j. If →c is a vector such that →a⋅→c=∣∣→c∣∣, ∣∣→c−→a∣∣=2√2 and the angle between (→a×→b) and →c is π6, then the value of ∣∣∣(→a×→b)×→c∣∣∣ is