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Worked solution

Give a direction vector the required magnitude

Find the vector p\mathbf p given that ∣p∣=6|\mathbf p|=6 and p\mathbf p is in the direction of 2i−3j+6k2\mathbf i-3\mathbf j+6\mathbf k.

CJC · Discussion · Question 1

Answer

p=67(2i−3j+6k)\mathbf p=\dfrac 67(2\mathbf i-3\mathbf j+6\mathbf k).

Normalise the given direction

The direction vector v=(2−36)\mathbf v=\begin{pmatrix}2\\-3\\6\end{pmatrix} has magnitude 22+(−3)2+62=7\sqrt{2^2+(-3)^2+6^2}=7. Its unit vector is p^=17(2−36)\widehat{\mathbf p}=\frac 17\begin{pmatrix}2\\-3\\6\end{pmatrix}.

How did this part go?

Scale to magnitude 6

The phrase “in the direction of” fixes the positive orientation. Thus

p=∣p∣p^=67(2−36).\mathbf p=|\mathbf p|\widehat{\mathbf p}=\frac 67\begin{pmatrix}2\\-3\\6\end{pmatrix}.

Its magnitude is 6 and it is a positive multiple of the supplied direction.

How did this part go?

Common mistakes

A negative multiple would have the opposite direction. Magnitude bars distinguish the scalar 6 from the vector itself.

Solution notes

The source visibly has |p|=6; extraction drops the bars. Restored from the page image, not treated as a source mathematical error.

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