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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Mathematics
Chapter
Vector Algebra

JEE Advanced 2011 · Paper 1 · Mathematics · Vector Algebra

Let a= i + j + k , b= i - j + k and c= i - j - k be three vectors....

Let a=i^+j^+k^,b=i^−j^+k^\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} and c=i^−j^−k^\mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} be three vectors. A vector v\mathbf{v} in the plane of a\mathbf{a} and b\mathbf{b} whose projection of c\mathbf{c} is 13\frac{1}{\sqrt{3}}, is given by

Options

  1. Ai^−3j^+3k^\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}
  2. B−3i^−3j^−k^-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}
  3. C3i^−j^+3k^3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}
  4. Di^+3k^−3k^\hat{\mathbf{i}}+3 \hat{\mathbf{k}}-3 \hat{\mathbf{k}}

Correct answer

C. 3i^−j^+3k^3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}

Step-by-step solution

Let v=a+λbv=\mathbf{v}=\mathbf{a}+\lambda \mathbf{b} \mathbf{v}=(1+\lambda) \hat{i\mathbf{i}}+(1-\lambda) \hat{j\mathbf{j}}+(1+\lambda) \hat{k\mathbf{k}} Projection of v\mathbf{v} on c=13\mathbf{c}=\frac{1}{\sqrt{3}}⇒v⋅c∣c∣=13⇒(1+λ)−(1−λ)−(1+λ)3=13⇒1+λ−1+λ−1−λ=1⇒λ−1=1⇒λ=2∴v=3i^−j^+3k^\begin{array}{lc} \Rightarrow& \frac{\mathbf{v} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}}\\ \Rightarrow & \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}}\\ \Rightarrow & 1+\lambda-1+\lambda-1-\lambda=1 \\ \Rightarrow & \lambda-1=1 \Rightarrow \lambda=2 \\ \therefore & \mathbf{v}=3 \hat{ \mathbf{i} }-\hat{ \mathbf{j} }+3 \hat{ \mathbf{k} } \end{array}
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