If a -k- b -k- and c is a unit vector perpendicular to the vector a and coplanar with a and b- then direction cosines of avector which is perpendicular to both a and c areA- 1-2- 1-2- 0B- 0- 1-2- 1-2C- -1-2- 0- 1-2D- 1-2- 0- 1-2

The correct option is B (0,1√(2),1√(2))cis coplanar with nbsp;a and nbsp;b ∴ nbsp;c=xa+yb =(x+y)î+(x y)ĵ+(y x)k̂ ⋯(1) ∵ |c|=1 ⇒ 3x^2+3y^2 2xy=1 ⋯(2) ...

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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 a unit vector perpendicular to the vector $\to a$ and coplanar with $\to a$ and $\to b$ , then direction cosines of a vector which is perpendicular to both $\to a$ and $\to c$ are

(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

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(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})

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(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)

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(- \frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})

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Solution

The correct option is A $(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
$\to c$ is coplanar with $\to a$ and $\to b$
$∴ \to c = x \to a + y \to b = (x + y)^i + (x - y)^j + (y - x)^k \dots (1)$
$∵ | \to c | = 1 \Rightarrow 3 x^{2} + 3 y^{2} - 2 x y = 1 \dots (2)$
$∵ \to c \cdot \to a = 0 ∴ x + y + x - y + x - y = 0 \Rightarrow 3 x = y$
from $(2)$
$x = \pm \frac{1}{2 \sqrt{6}}; y = \pm \frac{3}{2 \sqrt{6}}$
from $(1)$
$\to c = \pm (\frac{2}{\sqrt{6}}, - \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$

So,
$\to d = \pm ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k \frac{2}{\sqrt{6}} & - \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} 1 & 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \Rightarrow \to d = \pm (0, \frac{3}{\sqrt{6}}, \frac{3}{\sqrt{6}})$
$∴ unit vector^d = \frac{\to d}{| \to d |} = \pm (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

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