Correct answer: D. (i) (b), (ii) (a), (iii) (c), (iv) (d) (i) Solved the two equations, say, i.e. x+y=|a| and a x-y=1 , we get x=(|a|+1)/(a+1) 0 and y=(|a|-1)/(a+1) 0 when a+1 0 ; we get a 1 a_0=1 (ii) We have, a=α i +β j +γ k a · k =γ Now, k ×( k × a)=( k · a)-( k · k ) a=γ k -(α i +β j +γ k ) α i +β j =0 α=β=0 Also, α+β+γ=2 γ=2 (iv) A B C+ A B ≤ A B+ A B= (A-B) array lc & (A-B) ≥ 1 (A-B)=1 & C=1 array

JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2006 · Mathematics · Vector Algebra

Match the following

Options

A(i) (a), (ii) (b), (iii) (d), (iv) (c)
B(i) (c), (ii) (a), (iii) (b), (iv) (d)
C(i) (a), (ii) (d), (iii) (b), (iv) (c)
D(i) (b), (ii) (a), (iii) (c), (iv) (d)

Correct answer

D. (i) (b), (ii) (a), (iii) (c), (iv) (d)

Step-by-step solution

(i) Solved the two equations, say, i.e.

x+y=|a|

and

a x-y=1

, we get

x=\frac{|a|+1}{a+1} \gt 0

and

y=\frac{|a|-1}{a+1} \gt 0

when

a+1 \gt 0

; we get

a \gt 1\therefore \quad a_0=1

(ii) We have,

\mathbf{a}=\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}}\Rightarrow \quad \mathbf{a} \cdot \hat{\mathbf{k}}=\gamma

Now,

\hat{\mathbf{k}} \times(\hat{\mathbf{k}} \times \mathbf{a})=(\hat{\mathbf{k}} \cdot \mathbf{a})-(\hat{\mathbf{k}} \cdot \hat{\mathbf{k}}) \mathbf{a}=\gamma \hat{\mathbf{k}}-(\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}})\Rightarrow \quad \alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}=0 \Rightarrow \alpha=\beta=0

Also,

\quad \alpha+\beta+\gamma=2 \Rightarrow \gamma=2

(iv)

\sin A \sin B \sin C+\cos A \cos B \leq \sin A \sin B+\cos A \cos B=\cos (A-B)

\begin{array}{lc} \Rightarrow& \cos (A-B) \geq 1 \Rightarrow \cos (A-B)=1 \\ \Rightarrow & \sin C=1 \end{array}

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