Exams › JEE Main › Maths
Let a, b, c be three non-zero vectors, no two of which are collinear. If a + 3b is parallel to c and b + 2c is parallel to a, then the vector a + 3b + 6c is:
- a
- c
- 0
- a + c
Correct answer: 0
Solution
The conditions given imply that the vectors can be expressed in terms of each other, leading to a situation where the combination of these vectors results in the zero vector. Specifically, since both expressions are parallel to different vectors, their linear combinations ultimately cancel each other out, resulting in the vector being zero.
Related JEE Main Maths questions
- If ((â × b̂) × (ĉ × d̂)) · (â × d̂) = 0, which statement must always hold?
- For four points A, B, C and D in space, if the dot product of vectors AB and CD is given by k times [|AD|² + |BC|² - |AC|² - |BD|²], then what is the value of k?
- Three vectors →a, →b and →c have magnitudes |→a| = 4, |→b| = 4 and |→c| = 2. If →a is orthogonal to (→b + →c), →b is orthogonal to (→c + →a), and →c is orthogonal to (→a + →b), then the magnitude of (→a + →b + →c) is:
- The position vectors of the vertices A, B and C of triangle ABC are 4î + 7ĵ + 8k̂, 2î + 3ĵ + 4k̂ and 2î + 5ĵ + 7k̂ respectively. The position vector of the point where the internal bisector of angle A intersects BC is
- Let a=a₁î+a₂ĵ+a₃k̂, b=b₁î+b₂ĵ+b₃k̂, and c=c₁î+c₂ĵ+c₃k̂ be three non-zero vectors. Suppose c is a unit vector perpendicular to both a and b. If the angle between a and b is π/6, then the value of |[a₁, a₂, a₃; b₁, b₂, b₃; c₁, c₂, c₃] | is
- Let a, b, and c be three vectors that are not in the same plane. Define p = (b × c)/([abc]), q = (c × a)/([abc]), and r = (a × b)/([abc]). Then the value of (a+b)·p + (b+c)·q + (c+a)·r is
⚔️ Practice JEE Main Maths free + battle 1v1 →