Question

The direction cosines of the line which is perpendicular to the lines with direction ratios 1,-2,-2 and 0, 2, 1 are:

• A. \( \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \)
• B. \( -\frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \)
• C. \( \frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \)
• D. \( \frac{2}{3}, \frac{1}{3}, \frac{2}{3} \)

Hint:

When calculating direction cosines for a line perpendicular to two given lines, always start by finding the cross product of the direction ratios of the lines. Once you have the resulting vector, normalize it by dividing each component by the magnitude of the vector. This will give you the direction cosines. Remember that the cross product is a fundamental operation when dealing with perpendicular vectors in 3D space.

Answer 1

The Correct Option is A

To find the direction cosines of a line perpendicular to the given lines, we need to find a vector that is perpendicular to both vectors represented by the direction ratios (1, -2, -2) and (0, 2, 1). We achieve this by computing the cross product of these two vectors:  

\( \mathbf{a} = (1, -2, -2) \) 
\( \mathbf{b} = (0, 2, 1) \) 

The cross product \( \mathbf{a} \times \mathbf{b} \) is given by:
\(\mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -2 \\ 0 & 2 & 1 \end{vmatrix}\)
\(\mathbf{c} = \mathbf{i}((-2)(1) - (-2)(2)) - \mathbf{j}((1)(1) - (-2)(0)) + \mathbf{k}((1)(2) - (-2)(0))\)
\(\mathbf{c} = \mathbf{i}(-2 + 4) - \mathbf{j}(1) + \mathbf{k}(2)\)
\(\mathbf{c} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)

Thus, the direction ratios of the perpendicular line are (2, -1, 2).

Now, we convert these direction ratios to direction cosines by dividing each value by the magnitude of the vector.

Magnitude of vector: \(\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)

So, the direction cosines are:
\(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\)

Thus, the correct option is \( \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \).

Answer 2

Use the concept that the direction cosines of a line perpendicular to two given lines can be found by taking the cross product of the direction ratios.

We are given the direction ratios of two lines: \( \langle 1, -2, -2 \rangle \) and \( \langle 0, 2, 1 \rangle \). To find the direction cosines of the line perpendicular to these two lines, we compute the cross product of these two vectors.

Step 1: Compute the cross product of the vectors:

The cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:

\[ \mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1 \rangle \]

Substituting the values \( \langle 1, -2, -2 \rangle \) for \( \mathbf{a} \) and \( \langle 0, 2, 1 \rangle \) for \( \mathbf{b} \), we get:

\[ \mathbf{a} \times \mathbf{b} = \langle (-2)(1) - (-2)(2), (-2)(0) - (1)(1), (1)(2) - (-2)(0) \rangle \]

Simplifying the components:

\[ \mathbf{a} \times \mathbf{b} = \langle -2 + 4, 0 - 1, 2 - 0 \rangle = \langle 2, -1, 2 \rangle \]

Step 2: Normalize the resulting vector to get the direction cosines:

The direction cosines are obtained by normalizing the vector \( \langle 2, -1, 2 \rangle \). To normalize, we divide each component of the vector by its magnitude.

The magnitude of \( \langle 2, -1, 2 \rangle \) is:

\[ |\mathbf{v}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Now, divide each component by the magnitude:

\[ \text{Direction cosines} = \left\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \right\rangle \]

Conclusion:

The direction cosines of the line perpendicular to the two given lines are \( \left\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \right\rangle \). This confirms that option (1) is the correct answer.