If a -k- b -k- and c is unit vector perpendicular to the vector a and coplanar with a and b then a unit vector d perpendicular toboth a and c- isA- 1-62 -k-B- 1-2-k-C- 1-2-D- 1-2-k-

The correct option is B 1/√(2)(ĵ+k̂)We have, c⃗=±a⃗× (a⃗×b⃗)/|a⃗×(a⃗×b⃗)| Now, a⃗× (a⃗×b⃗)=(a⃗.b⃗)a⃗ (a⃗.a⃗)b⃗ = î ĵ+k̂ 3 (î ĵ+k̂)= 4î+2ĵ 2k̂ ∴c⃗=±1/√(6)(2î ĵ+ ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

If a =î+ĵ-k̂,...

Question

If $\to a =^i +^j -^k, \to b =^i -^j +^k$ , and $\to c$ is unit vector perpendicular to the vector $\to a$ and coplanar with $\to a$ and $\to b$ then a unit vector $\to d$ perpendicular to both $\to a$ and $\to c$ , is

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

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

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

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

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Solution

The correct option is B $\frac{1}{\sqrt{2}} (^j +^k)$
We have, $\to c = \pm \frac{\to a \times (\to a \times \to b)}{| \to a \times (\to a \times \to b) |}$
Now, $\to a \times (\to a \times \to b) = (\to a . \to b) \to a - (\to a . \to a) \to b$
$= -^i -^j +^k - 3 (^i -^j +^k) = - 4^i + 2^j - 2^k ∴ \to c = \pm \frac{1}{\sqrt{6}} (2^i -^j +^k)$
Since, $\to d$ is a unit vector perpendicular to both $\to a$ and $\to c$
Therefore, $\to d = \pm \frac{\to a \times \to c}{| \to a \times \to c |}$
Now, $\to a \times \to c = \pm \frac{1}{\sqrt{6}} ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 1 & - 1 2 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \pm \frac{1}{\sqrt{6}} (- 3^j - 3^k)$
$∴ \to d = \pm \frac{1}{\sqrt{2}} (-^j -^k) = \pm \frac{1}{\sqrt{2}} (^j +^k)$

Suggest Corrections

If →a=^i+^j−^k,→b=^i−^j+^k, and →c is unit vector perpendicular to the vector →a and coplanar with →a and →bthen a unit vector →d perpendicular to both →a and →c, is

If $\to a =^i +^j -^k, \to b =^i -^j +^k$ , and $\to c$ is unit vector perpendicular to the vector $\to a$ and coplanar with $\to a$ and $\to b$ then a unit vector $\to d$ perpendicular to both $\to a$ and $\to c$ , is