A line passes through (2, –1, 3) and is perpendicular to the line… - Vidyadip

Question

A line passes through (2, –1, 3) and is perpendicular to the lines$\overrightarrow{\text{r}}= (\hat{\text{i}} + \hat{\text{j}} - \hat{\text{k}}) + \lambda(2 \hat{\text{i}} - 2\hat{\text{j}} + \hat{\text{k}})\text{ and }\overrightarrow{\text{r}} = (2\hat{\text{i}} -\hat{\text{j}} - 3\hat{\text{k}}) + \mu(\hat{\text{i}} + 2 \hat{\text{j}} + 2\hat{\text{k}}).$Obtain its equation in vector and cartesian form.

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Answer

The direction perpendicular to the given lines is given by
$(2\hat{\text{i}} - 2 \hat{\text{j}} +\hat{\text{k}})\times(\hat{\text{i}} + 2\hat{\text{j}} + 2\hat{\text{k}})$
$= \begin{bmatrix} \hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}} \$0.3em] 2 & -2 & 1 \$0.3em] 1 & 2& 2 \end{bmatrix} = -6\hat{\text{i}} - 3\hat{\text{j}} + 6\hat{\text{k}}\text{ or }2\hat{\text{i}} +\hat{\text{j}} - 2\hat{\text{k}}$
$\therefore$ Vector equation of required line is
$\overrightarrow{\text{r}} = (2\hat{\text{i}} -\hat{\text{j}} + 3 \hat{\text{k}}) + \lambda(2\hat{\text{i}} + \hat{\text{j}} - 2 \hat{\text{k}})$
and the cartesian form is
$\frac{\text{x} - 2}{2} = \frac{\text{y} + 1 }{1} =\frac{\text{z} - 3}{-2}.$

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