Chapter 6 Line and Plane Ex 6.1 Solutions

The vector equation of the line passing through A () and parallel to the vector is = + λ, where λ is a scalar. ∴ the vector equation of the line passing through the point having position vector and parallel to the vector is .

The vector equation of the line passing through the A () and B() is , λ is a scalar ∴ the vector equation of the line passing through the points having position vectors and is

Let A be the point whose position vector is . Let be the vector parallel to the line having direction ratios -3, 4, 2 Then, = The vector equation of the line passing through A () and parallel to is , where λ is a scalar. ∴ the required vector equation of the line is .

Since the line is perpendicular to the vector and , it is parallel to . The vector equation of the line passing through A () and parallel to is , where λ is a scalar. Here, = Hence, the vector equation of the required line is

Let A be point having position vector = The required line is parallel to the line The vector equation of the line passing through A() and parallel to is = + λ where λ is a scalar. ∴ the required vector equation of the line is .

The cartesian equations of the line passing through (x1, y1, z1) 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

The cartesian equations of the line passing through the points (x1, y1, z1) and (x2, y2, z2) are Here, (x1, y1, z1) = (2, 2, 1) and (x2, y2, z2) = (1, 3, 0) ∴ the required cartesian equations are

We find the cartesian equations of the line AB. The cartesian equations of the line passing through the points (x1, y1, z1) and (x2, y2, z2) are = = Here, (x1, y1, z1) = (-2, 3, 4) and (x2, y2, z2) = (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.

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).

The vector perpendicular to the vectors and is given by Since the required line is perpendicular to the given lines, it is perpendicular to both and . ∴ it is parallel to The equation of the line passing through A() and parallel to is , where λ is a scalar. Here, = ∴ the equation of the required line is

The equation of the line is The coordinates of the origin O are (0, 0, 0) ∴ coordinates of the origin O satisfy the equation of the line. Hence, the line passes through the origin.