Question

Let \( P(\alpha, \beta, \gamma) \) be the point on the line \[ \frac{x-1}{2} = \frac{y+1}{-3} = z \] at a distance \( 4\sqrt{14} \) from the point \( (1,-1,0) \) and nearer to the origin. Then the shortest distance between the lines \[ \frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3} \quad \text{and} \quad \frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1} \] is equal to

• A. \( 7\sqrt{\frac{5}{4}} \)
• B. \( 4\sqrt{\frac{5}{7}} \)
• C. \( 2\sqrt{\frac{7}{4}} \)
• D. \( 4\sqrt{\frac{7}{5}} \)

Hint:

For shortest distance between two skew lines, always use the vector cross product formula involving direction vectors and a point-to-point joining vector.

Answer 1

The Correct Option is D

To solve the problem, we need to first identify the point \( P(\alpha, \beta, \gamma) \) on the given line that is at a distance \( 4\sqrt{14} \) from the point \( (1, -1, 0) \), and is nearer to the origin.

The line is given by the equation:

\[\frac{x-1}{2} = \frac{y+1}{-3} = z = t\]

From this parameterization, we can write the coordinates of any point on the line as:

\[(x, y, z) = (1 + 2t, -1 - 3t, t)\]

We want this point to be at a distance of \( 4\sqrt{14} \) from \( (1, -1, 0) \):

\[\sqrt{(1 + 2t - 1)^2 + (-1 - 3t + 1)^2 + (t - 0)^2} = 4\sqrt{14}\]

Simplifying, we get:

\[\sqrt{4t^2 + 9t^2 + t^2} = 4\sqrt{14}\]

This reduces to:

\[\sqrt{14t^2} = 4\sqrt{14}\]

Simplifying further, we have:

\[\sqrt{14} \cdot |t| = 4\sqrt{14}\]

Leading to:

\[|t| = 4\]

Thus, \( t = 4 \) or \( t = -4 \). We need the point nearer to the origin.

For \( t = 4 \):

\[(x, y, z) = (1 + 2 \times 4, -1 - 3 \times 4, 4) = (9, -13, 4)\]

Distance from origin \( = \sqrt{9^2 + (-13)^2 + 4^2} = \sqrt{246} \).

For \( t = -4 \):

\[(x, y, z) = (1 - 8, -1 + 12, -4) = (-7, 11, -4)\]

Distance from origin \( = \sqrt{(-7)^2 + 11^2 + (-4)^2} = \sqrt{186} \).

Hence, the point closer to the origin is \((-7, 11, -4)\).

Now, we find the shortest distance between the two lines given by:

\[\frac{x + 7}{1} = \frac{y - 11}{2} = \frac{z + 4}{3} \quad \text{and} \quad \frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}\]

The direction vectors for these lines are \( \mathbf{a} = \langle 1, 2, 3 \rangle \) and \( \mathbf{b} = \langle 2, 1, 1 \rangle \). The vector connecting any point on the first line to any point on the second line (like \((-7, 11, -4)\) to \((-5, 10, 3)\)) is \(\langle 2, -1, 7 \rangle\).

The shortest distance \( D \) between two skew lines is given by:

\[D = \frac{|(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}|}{|\mathbf{a} \times \mathbf{b}|}\]

where \( \mathbf{c} = \langle 2, -1, 7 \rangle \).

Calculating the cross product:

\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \langle 1, 5, -3 \rangle\]

Dot product:

\[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \langle 1, 5, -3 \rangle \cdot \langle 2, -1, 7 \rangle = 2 - 5 - 21 = -24\]

Magnitude of the cross product:

\[|\mathbf{a} \times \mathbf{b}| = \sqrt{1^2 + 5^2 + (-3)^2} = \sqrt{35}\]

Thus, the shortest distance is:

\[D = \frac{| -24 |}{\sqrt{35}} = \frac{24}{\sqrt{35}} = 4\sqrt{\frac{7}{5}}\]

Therefore, the shortest distance is \(4\sqrt{\frac{7}{5}}\).