Start Learning Free

Before You Begin

Question #41633

Difficulty: 8/10 | 🟢 Free

The Question

Let z=x+iyz = x + iy be a complex number satisfying ∣z∣2=56|z|^2 = 56. A line LL has direction vector v⃗=xi^+yj^+ck^\vec{v} = x\hat{i} + y\hat{j} + c\hat{k} (where c>0c > 0). Consider a plane PP with normal vector n⃗=i^+j^+k^\vec{n} = \hat{i} + \hat{j} + \hat{k} and a vector u⃗=2i^−j^+k^\vec{u} = 2\hat{i} - \hat{j} + \hat{k}. If the vectors satisfy the relation (v⃗⋅n⃗)(v⃗×u⃗)=v⃗×(n⃗×v⃗)+(v⃗⋅n⃗)v⃗−244n⃗(\vec{v} \cdot \vec{n}) (\vec{v} \times \vec{u}) = \vec{v} \times (\vec{n} \times \vec{v}) + (\vec{v} \cdot \vec{n})\vec{v} - 244 \vec{n} then the value of c2c^2 is:

Options

  • A.

    188

  • B.

    144

  • C.

    244

  • D.

    132

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

Collinearity vs Coplanarity|Line lying in a plane|Geometry of Complex Inequalities|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of 3D vector geometry and complex number constraints, specifically targeting the 'Zero-Factor Cancellation Trap'.

Students often attempt to algebraically manipulate the cross products without realizing that taking the dot product with u⃗\vec{u} elegantly isolates the magnitude ∣v⃗∣2=244|\vec{v}|^2 = 244.

A common pitfall is dividing by variable vector terms or assuming v⃗⋅n⃗≠0\vec{v} \cdot \vec{n} \neq 0, which prunes the valid geometric solution where the line is parallel to the plane (v⃗⋅n⃗=0\vec{v} \cdot \vec{n} = 0).

Furthermore, the problem forces students to verify the zero-product property (v⃗⋅n⃗)(v⃗×u⃗)=0⃗(\vec{v} \cdot \vec{n})(\vec{v} \times \vec{u}) = \vec{0}; they must use the complex modulus constraint (∣z∣2=56|z|^2 = 56) to rigorously eliminate the collinearity case (v⃗×u⃗=0⃗\vec{v} \times \vec{u} = \vec{0}), proving that the line's direction vector is indeed orthogonal to the plane's normal.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptCollinearity vs Coplanarity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree