Vector Algebra
Vector Algebra
nta_pyq_2025_jan
Grade 12
Question:
Let a ^ be a unit vector perpendicular to the vectors b = ^ \to i - 2 j + 3k and c = 2 i + 3 j - k, and makes an angle ^ ^ ^ ^ ^ of cos -1 (- 1 3 ) with the vector ^i + ^j + k ^ . If ^ a makes an angle of \pi 3 with the vector ^i + \alpha^j + k ^ , then the value of \alpha is :
\sqrt6
-\sqrt6
-\sqrt3
\sqrt3
Step-by-Step Solution
Key Concept: Apply the core result for dot product, cross product and projections and simplify using the given constraints.
∣ ^i ^ j ^ k ∣ ∣ ∣ ^ ^ ^ (2) ∣1 ∣ -2 3 ∣ = i (-7) + 7 j + 7k ∣ ∣2 3 -1 ∣ ^ ^ ^ ^ ^ ^ (-7 i + 7 j + 7k) -i + j + k ^ = $\pm$ a = $\pm$( ) \sqrt7 2 2 2 \sqrt3 + 7 + 7 (-1 + 1 + 1) 1 Now, cos \theta = $\pm$ = $\pm$ \sqrt3 ⋅ \sqrt3 3 ^ ^ ^ -1 -(- i + j + k) -1 \Rightarrow cos ( ^ = ) \Rightarrow a 3 \sqrt3 ^ ^ ^ i - j - k ^ = a \sqrt3 \pi 1 - \alpha - 1 cos = 3 \sqrt3 ⋅ \sqrt\alpha2 + 2 1 -\alpha = \to \alpha < 0 2 \sqrt3 ⋅ \sqrt\alpha 2 + 2 2 2 3 (\alpha + 2) = 4\alpha 2 6 = \alpha \alpha = $\pm$\sqrt6 Clearly, \alpha = -\sqrt6
Correct Answer: 2