If a- - i - j - k- b- - 1 - j - k- c- is a unit vector such that c- - a- - 0- -c-a-b- - 0 then a unit vectors perpendicular to both a- and c- is

The correct option is A 1√(2) (j + k) Let c⃗=xî+yĵ+zk̂ [c⃗ a⃗ b⃗]=0x amp;y amp;z 1 amp;1 amp; 1 1 amp; 1 amp;1=0 y+z=0⋯ (1) ...

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Standard XII

Mathematics

Definition of Vector

If a⃗ = i +...

Question

If $\to a = i + j - k, \to b = 1 - j + k, \to c$ is a unit vector such that $\to c . \to a = 0, [\to c \to a \to b] = 0$ then a unit vectors 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 A $\frac{1}{\sqrt{2}} (j + k)$
Let $\to c = x^i + y^j + z^k$
$[\to c \to a \to b] = 0 ⟹ ∣ ∣ ∣ ∣ \begin{matrix} x & y & z 1 & 1 & - 1 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 ⟹ y + z = 0 \dots (1)$
$\to c . \to a = 0 ⟹ x + y - z = 0 \dots (2)$
Solving $(1)$ and $(2)$
$∴ \frac{x}{- 2} = \frac{y}{1} = \frac{z}{- 1} ⟹ x = - 2 y, z = - y$
$x^{2} + y^{2} + z^{2} = 1 ⟹ 4 y^{2} + y^{2} + y^{2} = 1 ⟹ y = \frac{1}{\sqrt{6}}$
So $\to c = \frac{1}{\sqrt{6}} (- 2^i +^j -^k)$
Unit vector parallel to $\to a, \to c$ is $\frac{\to a \times \to c}{| \to a \times \to c |} = \frac{1}{\sqrt{2}} (^j +^k)$

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