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Vector Component

Projection of...

Question

Projection of $\to A =^i -^j$ along $\to B =^k$ is

A

- 2

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B

- 1

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C

1

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D

0

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Solution

The correct option is D $0$
We know that the preojection of $\to A along \to B$ is given as
$\frac{\to A \cdot \to B}{| \to B |}$
$\Rightarrow \frac{(^i -^j) . (^k)}{\sqrt{1^{2}}} = 0$

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Similar questions

\to a =^i + 2^j +^k, \to b =^i -^j +^k, \to c =^i +^j -^k .

A vector coplanar to

\to a

\to b

has a projection along

\to c

\frac{1}{\sqrt{3}}

, then the vector is

\to a =^i + 2^j +^k, \to b =^i -^j +^k, \to c =^i +^j -^k .

A vector coplanar to

\to a

\to b

has a projection along

\to c

\frac{1}{\sqrt{3}}

, then the vector is

\to a =^i + 2^j +^k, \to b =^i -^j +^k

\to c =^i +^j -^k

.The vector in the plane of

\to a

\to b

, whose projection on C is

\to a =^i +^j +^k, \to b =^i +^j -^k

\to c =^i -^j +^k

\to d =^i -^j -^k

(\to a \times \to b) \times (\to c \times \to d) =

\to a =^i +^j +^k

\to b =^i -^j +^k

\to c =^i +^j -^k

\to a =^i -^j -^k

(\to a \times \to b) \times (\to c \times \to d)

is a vector orthogonal to both

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