Let a-b-c- be unit vectors such that a-b- and c- is equally inclined to a- and b- If c-xa-yb-za-b- - then

The correct option is A z^2=1 2x^2 We have #160; a.b = 0 b.c = #160;a.c, ⇒ x = y #160;since #160; b.a = 0 #160;and cross product ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

Let a⃗,b⃗,c...

Question

Let $\to a, \to b, \to c$ be unit vectors such that $\to a ⊥ \to b$ and $\to c$ is equally inclined to $\to a$ and $\to b$ . If $\to c = x \to a + y \to b + z (\to a \times \to b)$ , then

z^{2} = 1 - 2 x^{2}

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z^{2} = 1 - x^{2}

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x^{2} = 2 y^{2}

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z^{2} = x^{2} + y^{2}

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Solution

The correct option is A $z^{2} = 1 - 2 x^{2}$
We have
$¯ ¯ ¯ a . ¯ ¯ b = 0$
$¯ ¯ b . ¯ ¯ c = ¯ ¯ ¯ a . ¯ ¯ c, \Rightarrow x = y$ since $¯ ¯ b . ¯ ¯ ¯ a = 0$ and cross product is perpendicular to both the vectors.
Also given that $¯ ¯ c$ is a unit vector which implies that $¯ ¯ c . ¯ ¯ c = 1$
$\Rightarrow x^{2} + y^{2} + z^{2} | ¯ ¯ ¯ a \times ¯ ¯ b |^{2} = 1.$
$∴ 2 x^{2} + z^{2} = 1$
$∴ z^{2} = 1 - 2 x^{2}$

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Let →a,→b,→c be unit vectors such that →a⊥→b and →c is equally inclined to →a and →b. If →c=x→a+y→b+z(→a×→b) , then

The correct option is A z2=1−2x2We have ¯¯¯a.¯¯b=0¯¯b.¯¯c=¯¯¯a.¯¯c,⇒x=y since ¯¯b.¯¯¯a=0 and cross product is perpendicular to both the vectors.Also given that ¯¯c is a unit vector which implies that ¯¯c.¯¯c=1⇒x2+y2+z2|¯¯¯a×¯¯b|2=1.∴2x2+z2=1∴z2=1−2x2

Let $\to a, \to b, \to c$ be unit vectors such that $\to a ⊥ \to b$ and $\to c$ is equally inclined to $\to a$ and $\to b$ . If $\to c = x \to a + y \to b + z (\to a \times \to b)$ , then