Two unit vectors a and b are pependicular to each other- Another unit vector c is inclined at an angle - to both a and b- If c- xa-yb-za-b- then

Given that nbsp; a·b=0, |a|^2=|b|^2=|c|^2=1 a·c=cosα, b·c=cosα a×b=n̂sin 90^∘⇒ |a×b|^2=1 a· (a×b)=0 and b· (a×b)=0 c=xa+yb+z(a×b) Scalar multiplication of ...

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Standard XIII

Mathematics

Linear Combination of Vectors

Two unit vect...

Question

Two unit vectors $\to a$ and $\to b$ are pependicular to each other. Another unit vector $\to c$ is inclined at an angle $α$ to both $\to a$ and $\to b$ . If $\to c = x \to a + y \to b + z (\to a \times \to b)$ , then

x^{2} + y^{2} = 1

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x^{2} = y^{2}

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z^{2} = - cos 2 α

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x^{2} + y^{2} + z^{2} = 1

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Solution

The correct option is D $x^{2} + y^{2} + z^{2} = 1$
Given that
$\to a \cdot \to b = 0, | \to a |^{2} = | \to b |^{2} = | \to c |^{2} = 1$

$\to a \cdot \to c = cos α, \to b \cdot \to c = cos α$
$\to a \times \to b =^n sin 90^{\circ} \Rightarrow | \to a \times \to b |^{2} = 1$
$\to a \cdot (\to a \times \to b) = 0$
and $\to b \cdot (\to a \times \to b) = 0$

$\to c = x \to a + y \to b + z (\to a \times \to b)$
Scalar multiplication of $\to a, \to b$ with $\to c$ one by one, we get
$\to c \cdot \to a = x \Rightarrow x = cos α \to c \cdot \to b = y \Rightarrow y = cos α$

Since, $| \to c |^{2} = 1 ∴ | x \to a + y \to b + z (\to a \times \to b) |^{2} = 1 \Rightarrow x^{2} \cdot 1 + y^{2} \cdot 1 + z^{2} \cdot 1 + 2 x y (0) + 2 y z (0) + 2 x z (0) = 1 \Rightarrow z^{2} + {cos}^{2} α + {cos}^{2} α = 1 \Rightarrow z^{2} = - (2 {cos}^{2} α - 1) \Rightarrow z^{2} = - cos 2 α$

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Two unit vectors →a and →b are pependicular to each other. Another unit vector →c is inclined at an angle α to both →a and →b. If →c=x→a+y→b+z(→a×→b), then

Two unit vectors $\to a$ and $\to b$ are pependicular to each other. Another unit vector $\to c$ is inclined at an angle $α$ to both $\to a$ and $\to b$ . If $\to c = x \to a + y \to b + z (\to a \times \to b)$ , then