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Question #40825

Difficulty: 8/10 | 🟢 Free

The Question

Let LL be the normal line to the parabola y2=4xy^2 = 4x at the point P(t2,2t)P(t^2, 2t), where t>0t > 0. Let u⃗\vec{u} be the vector parallel to LL such that u⃗⋅i^=1\vec{u} \cdot \hat{i} = 1. Let w⃗\vec{w} be the orthogonal projection of u⃗\vec{u} onto the plane x+y+z=0x + y + z = 0. If the volume of the tetrahedron formed by the origin OO and the points with position vectors a⃗=i^+4j^+k^\vec{a} = \hat{i} + 4\hat{j} + \hat{k}, b⃗=2i^+j^+k^\vec{b} = 2\hat{i} + \hat{j} + \hat{k}, and w⃗\vec{w} is zero, find the length of the segment OPOP.

Options

  • A.

    424\sqrt{2}

  • B.

    3133\sqrt{13}

  • C.

    55

  • D.

    252\sqrt{5}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Normal to Conic|Orthogonal projection of a line|Scalar Triple Product Volume Logic|Coplanarity Condition

Why This Question Is Tricky

Common Mistakes

This question tests the integration of 2D conic normals with 3D vector projections and scalar triple products.

The primary trap is the 'Coplanarity Condition' illusion: students often assume the normal vector u⃗\vec{u} remains in the xyxy-plane and directly apply [a⃗b⃗u⃗]=0[\vec{a} \vec{b} \vec{u}] = 0, ignoring the orthogonal projection onto the 3D plane.

This false assumption leads to t=3t=3 and the trap answer 3133\sqrt{13}.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterConic Sections - Class 11
ConceptNormal to Conic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree