Question #420ec

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Answer 1 · 2017-08-14T17:07:41

Given

we are to find $\to m and \to n$ $vecm and vecn$ satisfying conditions

By condition (b) $\to n$ $vecn$ is parallel to $\to p$ $vec p$

So

We can write
$\to n = α \cdot \to p$ $vecn=alpha*vecp$ ,where $α$ $alpha$ is a constant.

$\to n = α \to p = α ˆ i + 2 α ˆ j - 2 α ˆ k$ $vecn=alphavecp=alphahati+2alphahatj-2alphahatk$

By condition (c)

$\to m + \to n = \to q$ $vecm+vecn=vecq$

$\Rightarrow \to m = \to q - \to n$ $=>vecm=vecq-vecn$

$\Rightarrow \to m = (2 - α) ˆ i - (1 + 2 α) ˆ j + (2 + 2 α) ˆ k$ $=>vecm=(2-alpha)hati-(1+2alpha)hatj+(2+2alpha)hatk$

Now by condition (a) $\to m$ $vecm$ is perpendicular to $\to p$ $vec p$

So $\to m \cdot \to p = 0$ $vecm*vecp=0$

$\Rightarrow 1 \times (2 - α) - 2 (1 + 2 α) - 2 (2 + 2 α) = 0$ $=>1xx(2-alpha)-2(1+2alpha)-2(2+2alpha)=0$

$\Rightarrow 2 - α - 2 - 4 α - 4 - 4 α = 0$ $=>2-alpha-2-4alpha-4-4alpha=0$

$\Rightarrow α = - \frac{4}{9}$ $=>alpha=-4/9$

Now
$\to n = α ˆ i + 2 α ˆ j - 2 α ˆ k$ $vecn=alphahati+2alphahatj-2alphahatk$

$\Rightarrow \to n = - \frac{4}{9} ˆ i - \frac{8}{9} ˆ j + \frac{8}{9} ˆ k$ $=>vecn=-4/9hati-8/9hatj+8/9hatk$

and

$\to m = (2 - α) ˆ i - (1 + 2 α) ˆ j + (2 + 2 α) ˆ k$ $vecm=(2-alpha)hati-(1+2alpha)hatj+(2+2alpha)hatk$

$\Rightarrow \to m = (2 + \frac{4}{9}) ˆ i - (1 - \frac{8}{9}) ˆ j + (2 - \frac{8}{9}) ˆ k$ $=>vecm=(2+4/9)hati-(1-8/9)hatj+(2-8/9)hatk$

$\Rightarrow \to m = \frac{22}{9} ˆ i - \frac{1}{9} ˆ j + \frac{10}{9} ˆ k$ $=>vecm=22/9hati-1/9hatj+10/9hatk$