Question

Find the co-ordinate of a point on the line $\vec r = (\hat i - \hat j) + \lambda (2\hat i - 3\hat j + \hat k)$ at a distance $4\sqrt{14}$ from the point (1, –1, 0).

• A. (9, -13, 4)
• B. (-7, 11, -4)
• C. (1, -1, 0)
• D. (2, -3, 1)

Answer 1

Answer: (A), (B)

The co-ordinates of the points are $(9, -13, 4)$ and $(-7, 11, -4)$.

Answer 2

The equation of the line is given by $\vec r = (1\hat i - 1\hat j + 0\hat k) + \lambda (2\hat i - 3\hat j + 1\hat k)$. A general point $Q$ on this line has coordinates $(1 + 2\lambda, -1 - 3\lambda, \lambda)$. The given point is $P = (1, -1, 0)$. The square of the distance $PQ^2$ is: $PQ^2 = ((1 + 2\lambda) - 1)^2 + ((-1 - 3\lambda) - (-1))^2 + (\lambda - 0)^2$ $PQ^2 = (2\lambda)^2 + (-3\lambda)^2 + (\lambda)^2 = 4\lambda^2 + 9\lambda^2 + \lambda^2 = 14\lambda^2$. We are given that the distance $PQ = 4\sqrt{14}$, so $PQ^2 = (4\sqrt{14})^2 = 16 \times 14 = 224$. Equating the two expressions for $PQ^2$: $14\lambda^2 = 224$ $\lambda^2 = \frac{224}{14} = 16$ $\lambda = \pm 4$.

For $\lambda = 4$: $x = 1 + 2(4) = 9$ $y = -1 - 3(4) = -13$ $z = 4$ The point is $(9, -13, 4)$.

For $\lambda = -4$: $x = 1 + 2(-4) = -7$ $y = -1 - 3(-4) = 11$ $z = -4$ The point is $(-7, 11, -4)$.