Question

If the direction cosines of a line satisfy the relations \[ l-m+n=0 \] and \[ lm+mn-4nl=0, \] then the direction cosines of the line are:

• A. \[ \left( -\frac{1}{\sqrt6}, \frac{2}{\sqrt6}, \frac{1}{\sqrt6} \right) \]
• B. \[ \left( \frac{1}{\sqrt6}, -\frac{2}{\sqrt6}, \frac{1}{\sqrt6} \right) \]
• C. \[ \left( \frac{1}{\sqrt6}, \frac{2}{\sqrt6}, -\frac{1}{\sqrt6} \right) \]
• D. \[ \left( \frac{1}{\sqrt6}, \frac{2}{\sqrt6}, \frac{1}{\sqrt6} \right) \]

Hint:

For direction cosines, first use the given relations to find the ratio \(l:m:n\), then apply \[ l^2+m^2+n^2=1 \] to determine the actual values.

Answer 1

The Correct Option is D

Step 1: Express m in terms of l and n from the first relation.
l - m + n = 0 → m = l + n.

Step 2: Substitute into the second relation.
lm + mn - 4nl = 0 becomes l(l + n) + (l + n)n - 4ln = 0 → l² + ln + ln + n² - 4ln = 0 → l² - 2ln + n² = 0 → (l - n)² = 0 → l = n.

Step 3: Determine the ratio l : m : n.
With l = n and m = l + n = 2l, the ratio is l : m : n = 1 : 2 : 1.

Step 4: Normalize using the direction cosine condition.
l² + m² + n² = 1 → l² + (2l)² + l² = 1 → 6l² = 1 → l = ±1/√6. Taking the positive set gives (1/√6, 2/√6, 1/√6).

Step 5: Final conclusion.
The direction cosines are (1/√6, 2/√6, 1/√6).