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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Vector Algebra

NDA 2010 · Mathematics · Vector Algebra

A vector b is collinear with the vector a =(2,1,-1) and satisfies...

A vector b⃗\vec{b} is collinear with the vector a⃗=(2,1,−1)\vec{a}=(2,1,-1) and satisfies the condition a⃗⋅b⃗=3\vec{a} \cdot \vec{b}=3. What is b⃗\vec{b} equal to?

Options

  1. A(1,1/2,−1/2)(1,1 / 2,-1 / 2)
  2. B(2/3,1/3,−1/3)(2 / 3,1 / 3,-1 / 3)
  3. C(1/2,1/4,−1/4)(1 / 2,1 / 4,-1 / 4)
  4. D(1,1,0)(1,1,0)

Correct answer

A. (1,1/2,−1/2)(1,1 / 2,-1 / 2)

Step-by-step solution

b⃗=aˉλb⃗λ=2i^+j^−k^a⃗=2i^+j^−k^∴a⃗⋅b⃗=4λ+λ+λ=6λ=3or λ=12b⃗=(i^+12j^−12k^)=(1+12,−12)\begin{aligned} & \vec{b}=\bar{a} \lambda \\ & \vec{b} \lambda=2 \hat{i}+\hat{j}-\hat{k} \\ & \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \\ & \therefore \vec{a} \cdot \vec{b}=4 \lambda+\lambda+\lambda=6 \lambda=3 \\ & \text{or } \lambda=\frac{1}{2} \vec{b}=\left(\hat{i}+\frac{1}{2} \hat{j}-\frac{1}{2} \hat{k}\right) \\ & =\left(1+\frac{1}{2},-\frac{1}{2}\right)\end{aligned}
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