If {a} = {i} + λ {j} + 2{k}; {b} = {i} + {j} - {k} are orthogonal and |{a}| = |{b}| then ( λ , µ) =

Correct Answer is : (14,74)≤ft(1/4, 7/4)(41,47)

Correct Answer is : (14,74)≤ft(1/4, 7/4)(41,47)

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KCET 2018 Math Solved Paper

Question : 11 of 60

Marks: +1, -0

\vec{a} = \hat{i} + \lambda \hat{j} + 2\hat{k}

\vec{b} = \mu \hat{i} + \hat{j} - \hat{k}

\vec{a}

\vec{b}

( λ , µ) =

Solution:

Given vectors,

\vec{a} = \hat{i} + \lambda \hat{j} + 2\hat{k}

→ (1)

\vec{b} = \mu \hat{i} + \hat{j} - \hat{k}

→ (2)

For vectors to be orthogonal, we have

\vec{a} \cdot \vec{b} = 0

\Rightarrow (\hat{i} + \lambda \hat{j} + 2\hat{k}) \cdot (\mu \hat{i} + \hat{j} - \hat{k}) = 0

\Rightarrow \mu + \lambda - 2 = 0

\Rightarrow \lambda + \mu = 2

→ (3)

and

|\vec{a}| = |\vec{b}|

\Rightarrow \sqrt{1 + \lambda^{2} + 4} = \sqrt{\mu^{2} + 1 + 1}

\Rightarrow 1 + \lambda^{2} + 4 = \mu^{2} + 1 + 1

→ (4)

From Eqs. (3) and (4), we get

\lambda^{2} + 3 = \mu^{2} = (2 \;\; \lambda)^{2}

\Rightarrow \lambda^{2} + 3 = 4 + \lambda^{2} - 4\lambda

So,

\lambda = \frac{1}{4} \Rightarrow \mu = 2 - \frac{1}{4} = \frac{7}{4}

Hence

(\lambda, \mu) = \left(\frac{1}{4}, \frac{7}{4}\right)

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