PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.

Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. 

Reduce the equation \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.

The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. 

Find the equation of a plane which is at a distance of \[3\sqrt{3}\]  units from the origin and the normal to which is equally inclined to the coordinate axes.

Find the vector equation of the plane which is at a distance of \[\frac{6}{\sqrt{29}}\] from the origin and its normal vector from the origin is  \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} .\] Also, find its Cartesian form. 

Find the value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\]  is perpendicular to the plane 3x − y − 2z = 7.

Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line \[\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}\]

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j}  - 2 \hat{k}  + \lambda\left( \hat{i}  - \hat{j}  + \hat{k}  \right) .\]

Write the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right) = 14\]  in normal form.

Write a vector normal to the plane  \[\vec{r} = l \vec{b} + m \vec{c} .\]

Write the value of k for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\]  is perpendicular to the normal to the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  + 4 \hat{k}  \right) = 4 .\]

Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + \hat{j}  + 2 \hat{k}  \right) = 5 .\]

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .