Find the vector and the cartesian equation of the line that passes through $(-1, 2, 7)$ and is perpendicular to the lines $\vec{r} = 2\hat{i} + \hat{j} - 3\hat{k} + \lambda(\hat{i} + 2\hat{j} + 5\hat{k})$ and $\vec{r} = 3\hat{i} + 3\hat{j} - 7\hat{k} + \mu(3\hat{i} - 2\hat{j} + 5\hat{k})$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Vector: $\vec{r} = (-\hat{i} + 2\hat{j} + 7\hat{k}) + a(10\hat{i} + 5\hat{j} - 4\hat{k})$; Cartesian: $\frac{x+1}{10} = \frac{y-2}{5} = \frac{z-7}{-4}$ ##
Line perpendicular to the lines
$\vec{r} = 2\hat{i} + \hat{j} - 3\hat{k} + \lambda(\hat{i} + 2\hat{j} + 5\hat{k})$
and $\vec{r} = 3\hat{i} + 3\hat{j} - 7\hat{k} + \mu(3\hat{i} - 2\hat{j} + 5\hat{k})$
has a vector perpendicular it is given by
$\vec{b} = \vec{b_1} \times \vec{b_2}$
$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 5 \\ 3 & -2 & 5 \end{vmatrix} = 20\hat{i} + 10\hat{j} - 8\hat{k}$
$∴$ Equation of line in vector from is
$\vec{r} = -\hat{i} + 2\hat{j} + 7\hat{k} + a(10\hat{i} + 5\hat{j} - 4\hat{k})$
and the equation of a line in cartesian form is
$\frac{x+1}{10} = \frac{y-2}{5} = \frac{z-7}{-4}$