Two vectors a and b are expressed in terms of unit vectors as follows $a = 3 i + j + 2 k, b = 2 i - 2 j + 4 k .$
What is the unit vector perpendicular to each of the vectors ? Also determine the sine of the angle between the given vectors.

\frac{1}{\sqrt{14}} (2 i - j - 3 k), sin θ = \frac{3}{\sqrt{(14)}} .

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\frac{1}{\sqrt{14}} (- 2 i + j + 3 k), sin θ = \frac{3}{\sqrt{(14)}} .

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\frac{1}{\sqrt{3}} (i - j - k), sin θ = \frac{2}{\sqrt{(7)}} .

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\frac{1}{\sqrt{3}} (- i + j + k), sin θ = \frac{2}{\sqrt{(7)}} .

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Solution

The correct option is D $\frac{1}{\sqrt{3}} (i - j - k), sin θ = \frac{2}{\sqrt{(7)}} .$
The vector perpendicular to both vector $a$ and vector $b$ be vector $c$ . Vector $c$ can be calculated by the cross product of $a$ and $b$
By doing cross product of $a$ and $b$ we get $8 (i - j - k)$
Since $c$ is a unit vector, we get $c = (i - j - k) / (\sqrt{3})$
Let the angle between $a$ and $b$ be $z$ . So
$C o s (z) = (6 - 2 + 8) / (\sqrt{1} 4) (\sqrt{2} 4) = (12) / (\sqrt{3} 36) = (\sqrt{3}) / (\sqrt{7})$
Therefore $S i n (z) = 2 / (\sqrt{7})$

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