Find the equation of the plane passing through the point (−1, − 1, 2) and perpendicular to each of the following planes: 2x + 3y – 3z = 2   and   5x – 4y + z = 6

322.

Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line $\frac{\mathrm{x} + 3}{2} = \frac{\mathrm{y} - 3}{7} = \frac{\mathrm{z} - 2}{5}$

Find the value of λ so that the lines, $\frac{1 - \mathrm{x}}{3} = \frac{\mathrm{y} - 2}{2\mathrm{\lambda }} = \frac{\mathrm{z} - 3}{2} \mathrm{and} \frac{\mathrm{x} - 1}{3\mathrm{\lambda }} = \frac{\mathrm{y} - 1}{1} = \frac{6 - \mathrm{z}}{7}$ are perpendicular to each other.

Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

What is the cosine of the angle which the vector $\sqrt{2} \hat{\mathrm{i}} + \hat{\mathrm{j}} + \mathrm{k}$  makes with y-axis?

Write the vector equation of the following line:

$\frac{\mathrm{x} - 5}{3} = \frac{\mathrm{y} + 4}{7} = \frac{6 - \mathrm{z}}{2}$

Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, -1, 2) and parallel to the line  $\frac{\mathrm{x} - 4}{1} = \frac{\mathrm{y} + 3}{-4} = \frac{\mathrm{z} + 1}{7}$

Write the vector equations of the following lines and hence determine the distance between them:

$\frac{ \mathrm{x} -1}{2} = \frac{\mathrm{y} - 2}{3} = \frac{\mathrm{z} + 4}{6}; \frac{\mathrm{x} - 3}{4} = \frac{\mathrm{y} - 3}{6} = \frac{\mathrm{z} + 5}{12}$

Write the intercept cut off by the plane 2x + y – z = 5 on x-axis.

Find the angle between the following pair of lines:  

$\frac{- \mathrm{x} + 2}{- 2} = \frac{\mathrm{y} - 1}{7} = \frac{\mathrm{z} + 3}{- 3} \mathrm{and} \frac{\mathrm{x} + 2}{- 1} = \frac{2 \mathrm{y} - 8}{4} = \frac{\mathrm{z} - 5}{4}$

And check whether the lines are parallel or perpendicular.