**Question 1.**

**Question 2.**

**Question 3.**

**Question 4.Solution:**

**Question 5.**

**Question 6.Find the Cartesian equations of the line passing through A(-1, 2, 1) and having direction ratios 2, 3, 1.Solution:**

The cartesian equations of the line passing through (x

_{1}, y

_{1}, z

_{1}) and having direction ratios a, b, c are

∴ the cartesian equations of the line passing through the point (-1, 2, 1) and having direction ratios 2, 3, 1 are

**Question 7.Find the Cartesian equations of the line passing through A(2, 2, 1) and B(1, 3, 0).Solution:**

The cartesian equations of the line passing through the points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) are

Here, (x

_{1}, y

_{1}, z

_{1}) = (2, 2, 1) and (x

_{2}, y

_{2}, z

_{2}) = (1, 3, 0)

∴ the required cartesian equations are

**Question 8.A(-2, 3, 4), B(1, 1, 2) and C(4, -1, 0) are three points. Find the Cartesian equations of the line AB and show that points A, B, C are collinear.Solution:**

We find the cartesian equations of the line AB. The cartesian equations of the line passing through the points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) are

Here, (x

_{1}, y

_{1}, z

_{1}) = (-2, 3, 4) and (x

_{2}, y

_{2}, z

_{2}) = (4, -1, 0)

∴ the required cartesian equations of the line AB are

∴ coordinates of C satisfy the equations of the line AB.

∴ C lies on the line passing through A and B.

Hence, A, B, C are collinear.

**Question 9.Solution:**

The equations of the lines are

From (1), x = -1 -10λ, y = -3 – 2, z = 4 + λ

∴ the coordinates of any point on the line (1) are

(-1 – 10λ, – 3 – λ, 4 + λ)

From (2), x = -10 – u, y = -1 – 3u, z = 1 + 4u

∴ the coordinates of any point on the line (2) are

(-10 – u, -1 – 3u, 1 + 4u)

Lines (1) and (2) intersect, if

(- 1 – 10λ, – 3 – λ, 4 + 2) = (- 10 – u, -1 – 3u, 1 + 4u)

∴ the equations -1 – 10λ = -10 – u, -3 – 2= – 1 – 3u

and 4 + λ = 1 + 4u are simultaneously true.

Solving the first two equations, we get, λ = 1 and u = 1. These values of λ and u satisfy the third equation also.

∴ the lines intersect.

Putting λ = 1 in (-1 – 10λ, -3 – 2, 4 + 2) or u = 1 in (-10 – u, -1 – 3u, 1 + 4u), we get

the point of intersection (-11, -4, 5).

**Question 10.Solution:**

The vector perpendicular to the vectors b¯ and c¯ is given by

∴ the equation of the required line is

**Question 11.**

∴ coordinates of the origin O satisfy the equation of the line.

Hence, the line passes through the origin.