Let a- - - -k- -b- - - -k- and c- be a unit vector perpendicular to a- and coplanar with a- and b- then c- is

The correct option is D 1 √( 6 )( 2î ĵ +k̂) Required unit vector =±a⃗×( a⃗×b⃗ #160; ) #160; | a⃗×( a⃗×b⃗ #160; ) #160; | #160; Now a⃗ ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

Let a⃗ =î +...

Question

Let $\to a =^i +^j -^k, \to b =^i -^j +^k$ and $\to c$ be a unit vector perpendicular to $\to a$ and coplanar with $\to a$ and $\to b$ , then $\to c$ is

\frac{1}{\sqrt{2}} (^j +^k)

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\frac{1}{\sqrt{2}} (^j -^k)

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\frac{1}{\sqrt{6}} (^i - 2^j +^k)

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\frac{1}{\sqrt{6}} (2^i -^j +^k)

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Solution

The correct option is D $\frac{1}{\sqrt{6}} (2^i -^j +^k)$
Required unit vector $= \pm \frac{\to a \times (\to a \times \to b)}{∣ ∣ \to a \times (\to a \times \to b) ∣ ∣}$
Now $\to a \times (\to a \times \to b) = (\to a . \to b) \to a - (\to a . \to a) \to b$
$\to a \times \to b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 1 & - 1 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 2^j - 2^k$
$∴ \to a \times (\to a \times \to b) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 1 & - 1 0 & - 2 & - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 4^i + 2^j - 2^k$
$∣ ∣ \to a \times (\to a \times \to b) ∣ ∣ = \sqrt{24} = 2 \sqrt{6}$
$∴$ Unit vector $= \pm \frac{\to a \times (\to a \times \to b)}{∣ ∣ \to a \times (\to a \times \to b) ∣ ∣} = \pm \frac{(2^i -^j +^k)}{\sqrt{6}}$
(taking positive for option)

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Similar questions

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Let →a=^i+^j−^k,→b=^i−^j+^k and →c be a unit vector perpendicular to →a and coplanar with →a and →b, then →c is

Let $\to a =^i +^j -^k, \to b =^i -^j +^k$ and $\to c$ be a unit vector perpendicular to $\to a$ and coplanar with $\to a$ and $\to b$ , then $\to c$ is