Question:
If the distance of point \(B(2,1,-3)\) from the line passing through the point \(A(4,-2,2)\), and parallel to the vector \(\overset{⃗}{c} = -4\hat{i}-6\hat{j}-2\hat{k}\) is \(x\), then \(x^4+x^2+541 =\)
If the distance of point \(B(2,1,-3)\) from the line passing through the point \(A(4,-2,2)\), and parallel to the vector \(\overset{⃗}{c} = -4\hat{i}-6\hat{j}-2\hat{k}\) is \(x\), then \(x^4+x^2+541 =\)
Show Hint
Move along the line direction until the plane is reached.
Updated On: Oct 1, 2026
- \(2026\)
- \(2025\)
- \(2024\)
- \(2023\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Parametric line:
The line through \((1, 0, -3)\) parallel to \((2, 3, -6)\) is \((1 + 2t,\ 3t,\ -3 - 6t)\).
Step 2: Meet the plane:
Substitute in \(x - y - z = 9\): \(1 + 2t - 3t + 3 + 6t = 9\), so \(4 + 5t = 9\) and \(t = 1\). The point is \((3, 3, -9)\).
Step 3: Distance:
Distance \(= |t|\sqrt{2^2 + 3^2 + 6^2} = 1\times7 = 7\) units.
Final Answer:
The distance is 7 units, option (B). \[ \boxed{7\text{ units}} \]
The line through \((1, 0, -3)\) parallel to \((2, 3, -6)\) is \((1 + 2t,\ 3t,\ -3 - 6t)\).
Step 2: Meet the plane:
Substitute in \(x - y - z = 9\): \(1 + 2t - 3t + 3 + 6t = 9\), so \(4 + 5t = 9\) and \(t = 1\). The point is \((3, 3, -9)\).
Step 3: Distance:
Distance \(= |t|\sqrt{2^2 + 3^2 + 6^2} = 1\times7 = 7\) units.
Final Answer:
The distance is 7 units, option (B). \[ \boxed{7\text{ units}} \]
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