A nonzero vector a is parallel to the...

MathematicsEquation of a PlaneJEE Advanced 1996Moderate

A nonzero vector $a$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i} + \hat{j}$ and the plane determined by the vectors $\hat{i} - \hat{j}, \hat{i} + \hat{k}$ . The angle between $a$ and the vector $\hat{i} - 2 \hat{j} + 2 \hat{k}$ is .........

Answer:

\frac{π}{4}

\frac{3 π}{4}

Visualized Solution (Hindi)

Visualizing the Planes

Identify the two planes that define the intersection line.
Plane 1 is determined by vectors $\hat{i}$ and $\hat{i} + \hat{j}$ .
Plane 2 is determined by vectors $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$ .

Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1

Equation of Plane 1:
Normal vector $n_{1} = \hat{i} \times (\hat{i} + \hat{j}) = \hat{k}$
Equation: $z = 0$

Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Equation of Plane 2:
Normal vector $n_{2} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{k}) = - \hat{i} - \hat{j} + \hat{k}$
Equation: $x + y - z = 0$

Line of Intersection

Direction of intersection line:
$v = n_{1} \times n_{2} = \hat{i} - \hat{j}$
Vector $a$ is parallel to $\hat{i} - \hat{j}$

Angle Setup

Angle between $a$ and $b = \hat{i} - 2 \hat{j} + 2 \hat{k}$ :
$cos θ = \pm \frac{a \cdot b}{∣ a ∣∣ b ∣}$

Calculation

Calculate dot product and magnitudes:
$a \cdot b = (1) (1) + (- 1) (- 2) + (0) (2) = 3$
$∣ a ∣ = 2$ , $∣ b ∣ = 3$
$cos θ = \pm \frac{3}{3 2} = \pm \frac{1}{2}$

Final Conclusion

Final angles:
$θ = \frac{π}{4}$ or $θ = \frac{3 π}{4}$
Key Takeaway: The line of intersection is perpendicular to both plane normals.
Next Challenge: What if the planes were parallel? Would an intersection line exist?

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