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Multiplication with Vectors

If with refer...

Question

If with reference to the right handed system of mutually perpendicular unit vector $^i,^j$ and $^k$ , $\to α = 3^i -^j, \to β = 2^i +^j - 3^k,$ then express $\to β$ in the form $\to β = {\to β}_{1} + {\to β}_{2}$ , where ${\to β}_{1}$ is parallel to $α$ and ${\to β}_{2}$ is perpendicular to $α$ .

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Solution

$\begin{matrix} \to α = 3^i -^j \to β = 2^i +^j - 3^k \end{matrix}$

Let,

$β_{1} = λ (3^i -^j)$
then,
$\to β =_{1} +_{2}$

$\to β_{2} = \to β - \to β_{1}$

$\begin{matrix} \to β_{2} = (2^i +^j - 3^k) - (3 λ^i - λ^j) \to β_{2} = (2 - 3 λ)^i + (1 + λ)^j - 3^k \end{matrix}$

Since, $_{2}$ is perpendicular to $α$
So,

$3 (2 - 3 λ) + (- 1) (1 + λ) + 0 (- 3) = 0$

$λ = \frac{1}{2}$

$β_{1} = \frac{1}{2} (3^i -^j) = \frac{3}{2}^i - \frac{1}{2}^j$

$β_{2} = \frac{1}{2}^i + \frac{3}{2}^j - 3^k$
Hence,

$\to β =_{1} +_{2}$
where,

$β_{1} = \frac{3}{2}^i - \frac{1}{2}^j$ and $β_{2} = \frac{1}{2}^i + \frac{3}{2}^j - 3^k$

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