Discussion
A unit vector parallel to the straight line \(\vec{r} = -(5 + 4s)\hat{i} + (7 - 2s)\hat{j} + (3 + 4s)\hat{k}\), where \(s\) is the parameter of the line, is
Answer: A
The unit vector is \(\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}\).
The direction of the line is the coefficient of the parameter \(s\) in each component:
\[ \begin{aligned} \vec{d} &= -4\hat{i} - 2\hat{j} + 4\hat{k} = 2(-2\hat{i} - \hat{j} + 2\hat{k}) \\ |-2\hat{i} - \hat{j} + 2\hat{k}| &= \sqrt{4 + 1 + 4} = 3 \end{aligned} \]
Dividing by the magnitude 3 gives the unit vector. Note the minus sign outside the bracket in the i-component: \(-(5 + 4s)\) contributes \(-4s\). The other options divide the same vector by 4, 5, 6 or 7, which do not give length 1.
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