Let -k- -k- and Y -k- be three vectors- A vector - in the plane of - and - whose projection on Y is1-3- is given by-A- -3 -3 k-B- -3 -3 k-C- -3 -3 k-D- -3 -3 k-

The correct option is C î + 3 ĵ + 3 k̂δ=α+ mβ = (î + ĵ + k̂) + m(î ĵ k̂) =(1+m)î+(1 m)ĵ+(1 m)k̂ δ·γ|γ| = 1√(3) ⇒ (1+m)+(1 m) (1 m)√(3) = 1√(3) ⇒ (1 ...

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Byju's Answer

Standard XII

Mathematics

Bisector of Angle between Two Vectors

Let α⃗=î+ĵ+k̂...

Question

Let $\to α =^i +^j +^k, \to β =^i -^j -^k$ and $\to γ = -^i +^j -^k$ be three vectors. A vector $\to δ,$ in the plane of $\to α$ and $\to β,$ whose projection on $\to γ$ is $\frac{1}{\sqrt{3}},$ is given by

-^i - 3^j - 3^k

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^i - 3^j - 3^k

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-^i + 3^j + 3^k

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^i + 3^j - 3^k

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Solution

The correct option is C $-^i + 3^j + 3^k$
$\to δ = \to α + m \to β = (^i +^j +^k) + m (^i -^j -^k)$
$= (1 + m)^i + (1 - m)^j + (1 - m)^k$

$\frac{\to δ \cdot \to γ}{| \to γ |} = \frac{1}{\sqrt{3}}$
$\Rightarrow \frac{- (1 + m) + (1 - m) - (1 - m)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$
$\Rightarrow - (1 + m) = 1$
$\Rightarrow m = - 2$

$∴ \to δ = -^i + 3^j + 3^k$

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