Exams › JEE Advanced › Maths

A non-zero vector a is parallel to the line of intersection of: Plane 1 (determined by vectors i and i+j) and Plane 2 (determined by vectors i-j and i-k). The possible angle(s) between a and the vector i - 2j + 2k is:

1. pi/3
2. pi/4
3. pi/6
4. 3*pi/4

Correct answer: pi/4

Solution

Normal to Plane 1: n1 = i x (i+j) = k. Normal to Plane 2: n2 = (i-j) x (i-k) = i+j+k. Direction of intersection: a || n1 x n2 = k x (i+j+k) = j - i = (-1, 1, 0). Angle with b = (1,-2,2): cos(theta) = [(-1)(1)+(1)(-2)] / (sqrt(2)*3) = -3/(3*sqrt(2)) = -1/sqrt(2). So theta = 3*pi/4 or pi/4 (acute).

Related JEE Advanced Maths questions

• Given that cos α ≠ 1, cos β ≠ 1, and cos γ ≠ 1, the vectors →a = i cos α + j cos β + k cos γ, →b = i + j cos β + k, and →c = i + j + k cos γ are
• Given the vectors →a = a→i + 2→j − 3→k, →b = →i + 2a→j − 2→k, and →c = 2→i − a→j + →k, if (→a × →b) × (→b × →c) × (→c × →a) equals →0, what is the value of a?
• In the triangle ΔPQR, the vectors →a, →b, and →c represent →QR, →RP, and →PQ respectively. If the magnitudes are |→a| = 12 and |→b| = 4√3, and the dot product →b ⋅ →c equals 24, which of the following statements is correct?
• Consider three vectors →x, →y, and →z, each having a magnitude of √2, with the angle between any two of them being π/3. If →a is a non-zero vector orthogonal to both →y and →z, and →b is a non-zero vector orthogonal to both →x and →z, which of the following is true?
• Let a, b, and c represent three unit vectors that are not in the same plane, with the angle between each pair being π/3. If the expression a × b + b × c equals pa + qb + rc, where p, q, and r are constants, what is the value of (p² + 2q² + r²)/q²?
• The vector a = (1, 3, sin 2α) forms an angle greater than 90° with the z-axis. Given that 2α is negative, and vectors b and c are perpendicular, b ⋅ c equals zero. The equation tan² α − tan α − 6 = 0 has solutions tan α = 3 or −2. If tan α = 3, then sin 2α = 2 tan α / (1 + tan² α) = 3/5, which is positive and contradicts the condition (2α < 0). Thus, tan α = −2. For this value, sin 2α = 2 tan α / (1 − tan² α) = 4/3, which is greater than zero. However, sin 2α must be negative, placing 2α in the third quadrant and α/2 in the first quadrant. The square root of sin(α/2) is valid, and α is given by (4n + 1)π − tan⁻¹ 2.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →