Find the projection of A - - - - k- along the vector B - 2- - - k-

The projection of vector nbsp;A in the direction of vector nbsp;B, nbsp; nbsp;P = | nbsp;A | cosθ A . B = |A ||B | cosθ |A | cosθ = A .B|B | P = A. B|B| ...

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Standard XII

Physics

Multiplication with Vectors

Find the proj...

Question

Find the projection of $\to A =^i +^j +^k$ along the vector $\to B = 2^i -^j +^k$

\frac{2}{\sqrt{6}}

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\frac{1}{\sqrt{6}}

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\frac{4}{\sqrt{6}}

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\frac{2}{\sqrt{3}}

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Solution

The correct option is A $\frac{2}{\sqrt{6}}$
The projection of vector $\to A$ in the direction of vector $\to B$ ,
$P = | \to A | cos θ$
$\to A . \to B = | \to A | | \to B | cos θ$
$| \to A | cos θ = \frac{\to A . \to B}{| \to B |}$
$P = \frac{\to A . \to B}{| \to B |}$
$P = \frac{(^i +^j +^k) . (2^i -^j +^k)}{\sqrt{2^{2} + (- 1)^{2} + 1^{2}}}$
$P = \frac{(2 - 1 + 1)}{\sqrt{6}} = \frac{2}{\sqrt{6}}$

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Find the projection of →A=^i+^j+^k along the vector →B=2^i−^j+^k

The correct option is A 2√6The projection of vector →A in the direction of vector →B, P=|→A|cosθ →A.→B=|→A||→B|cosθ |→A|cosθ=→A.→B|→B| P=→A.→B|→B| P=(^i+^j+^k).(2^i−^j+^k)√22+(−1)2+12 P=(2−1+1)√6=2√6

Find the projection of $\to A =^i +^j +^k$ along the vector $\to B = 2^i -^j +^k$