Let a and b be two unit vectors such that a-b - 0- For some - - - let c - x a - y b - a-b- If -c- - 2 and the vector c is inclined at the same angle - to both a and b- then the value of 8 cos2 - is -

C = x a + y b + (a×b) c· nbsp;a = x ⇒ |c||a| cosα nbsp; = x ⇒ x = 2cosα .............(i) c· nbsp;b = y ⇒ |c||b| cosα nbsp; = y ⇒ y = 2cosα ......... ...

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Byju's Answer

Standard XI

Mathematics

Dot Product of Two Vectors

Let a and b b...

Question

Let $\to a$ and $\to b$ be two unit vectors such that $\to a \cdot \to b = 0$ . For some $x, y \in R$ , let $\to c = x \to a + y \to b + (\to a \times \to b)$ . If $| \to c | = 2$ and the vector $\to c$ is inclined at the same angle $α$ to both $\to a$ and $\to b$ , then the value of $8 {cos}^{2} α$ is .

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Solution

$\to c = x \to a + y \to b + (\to a \times \to b)$
$\to c \cdot \to a = x \Rightarrow | \to c | | \to a | cos α = x \Rightarrow x = 2 cos α . . . . . . . . . . . . . (i)$
$\to c \cdot \to b = y \Rightarrow | \to c | | \to b | cos α = y \Rightarrow y = 2 cos α . . . . . . . . . . . . . (i i)$
$∴ x = y = 2 cos α$
We know $| \to a \times \to b | = 1$
$\to c \cdot \to c = x^{2} + y^{2} + 1^{2} \Rightarrow 4 = x^{2} + y^{2} + 1$
$\Rightarrow x^{2} + y^{2} = 3$ [ $∵ | \to c | = 2$ ]
Using equation $(i)$ and $(i i)$ ,
$8 {cos}^{2} α = 3$

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Let →a and →b be two unit vectors such that →a⋅→b=0. For some x,y∈R, let →c=x→a+y→b+(→a×→b). If |→c|=2 and the vector →c is inclined at the same angle α to both →a and →b, then the value of 8cos2α is .

→c=x→a+y→b+(→a×→b) →c⋅→a=x⇒|→c||→a|cosα=x⇒x=2cosα.............(i) →c⋅→b=y⇒|→c||→b|cosα=y⇒y=2cosα.............(ii) ∴x=y=2cosα We know |→a×→b|=1 →c⋅→c=x2+y2+12⇒4=x2+y2+1 ⇒x2+y2=3 [∵|→c|=2] Using equation (i) and (ii), 8cos2α=3