A-2-k-b-2-k- and a unit vector c- be coplanar- If c- is perpendicular to a- - then c-

The correct option is A #10; ±1√(2)( #160;j+#160;k) #10;Since, a⃗,b⃗,c⃗ are coplanar. ⇒c⃗=xa⃗+yb⃗ #160; #160;....(1)Since, c⃗⊥ ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

a⃗=2î+ĵ+k̂,b⃗...

Question

$\to a = 2^i +^j +^k, \to b =^i + 2^j -^k$ and a unit vector $\to c$ be coplanar. If $\to c$ is perpendicular to $\to a$ , then $\to c =$

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

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

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

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

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Solution

The correct option is A $\pm \frac{1}{\sqrt{2}} (-^j +^k)$
Since, $\to a, \to b, \to c$ are coplanar.
$\Rightarrow \to c = x \to a + y \to b$ ....(1)
Since, $\to c ⊥ \to a$
$\Rightarrow \to c . \to a = 0$
$\Rightarrow (x \to a + y \to b) . \to a = 0$
$\Rightarrow x {\to a}^{2} + y \to b . \to a = 0$
$\Rightarrow 6 x + 3 y = 0$
$\Rightarrow y = - 2 x$

So, by (1), we get
$\to c = x (\to a - 2 \to b)$
$\to c = x (- 3^i + 3^j)$
Unit vector $\to c = \pm \frac{1}{\sqrt{2}} (-^j +^k)$

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→a=2^i+^j+^k,→b=^i+2^j−^k and a unit vector →c be coplanar. If →c is perpendicular to →a , then →c=

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