Question:
Let \(\vec{a}, \vec{b}, \vec{c}\) be any three non-coplanar vectors. If the points \(\mathrm{A}(-2 \vec{a}+3 \vec{b}+5 \vec{c})\), \(\mathrm{B}(\vec{a}+2 \vec{b}+3 \vec{c})\) and \(\mathrm{C}(7 \vec{a}-\lambda \vec{c})\) are collinear, then \(\lambda\) is equal to :
Let \(\vec{a}, \vec{b}, \vec{c}\) be any three non-coplanar vectors. If the points \(\mathrm{A}(-2 \vec{a}+3 \vec{b}+5 \vec{c})\), \(\mathrm{B}(\vec{a}+2 \vec{b}+3 \vec{c})\) and \(\mathrm{C}(7 \vec{a}-\lambda \vec{c})\) are collinear, then \(\lambda\) is equal to :
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Collinearity means AB = k BC; compare the coefficients of the independent vectors a, b, c.
Updated On: Nov 12, 2026
- 1
- \(-1\)
- 2
- \(-2\)
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The Correct Option is A
Solution and Explanation
AB = B - A = 3a - b - 2c.
BC = C - B = 6a - 2b + (-λ - 3)c.
A, B, C are collinear, so AB = k BC. Because a, b, c are non-coplanar, coefficients must match: 3 = 6k, so k = 1/2; -1 = -2k holds.
For c: -2 = (1/2)(-λ - 3), so -λ - 3 = -4 and λ = 1.
Answer: (1) 1
BC = C - B = 6a - 2b + (-λ - 3)c.
A, B, C are collinear, so AB = k BC. Because a, b, c are non-coplanar, coefficients must match: 3 = 6k, so k = 1/2; -1 = -2k holds.
For c: -2 = (1/2)(-λ - 3), so -λ - 3 = -4 and λ = 1.
Answer: (1) 1
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