If {a}={i}+2{j}+{k}, {b}={i}-{j}+4{k} and {c}={i}+{j}+{k} are such that {a}+λ {b} is perpendicular to {c}, then the value of λ is

Correct Answer is : −1\;-1−1

Correct Answer is : −1\;-1−1

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KCET 2025 Math Paper

Question : 35 of 60

Marks: +1, -0

\vec{a}=\hat{i}+2\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+4\hat{k}

\vec{c}=\hat{i}+\hat{j}+\hat{k}

\vec{a}+\lambda \vec{b}

\vec{c}

\lambda

Solution:

Given:

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

and

\;\; \vec{b}=\hat{i}-\hat{j}+4\hat{k}

\;\vec{c}=\hat{i}+\hat{j}+\hat{k}

We need the expression

\vec{a}+\lambda \vec{b}

to be perpendicular to

\vec{c}

(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0

First, calculate

\vec{a}+\lambda \vec{b}

\vec{a}+\lambda \vec{b}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+4\hat{k})

Expanding this, we get:

\;=\hat{i}+2\hat{j}+\hat{k}+\lambda \hat{i}-\lambda \hat{j}+4\lambda \hat{k}

\;=(1+\lambda)\hat{i}+(2-\lambda)\hat{j}+(1+4\lambda)\hat{k}

Next, calculate the dot product with

\vec{c}

((1+\lambda)\hat{i}+(2-\lambda)\hat{j}+(1+4\lambda)\hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k})

Performing the dot product:

\;=(1+\lambda) \cdot 1+(2-\lambda) \cdot 1+(1+4\lambda) \cdot 1

\;=(1+\lambda)+(2-\lambda)+(1+4\lambda)

\;=1+\lambda+2-\lambda+1+4\lambda

\;=4+4\lambda

Set the expression to zero for perpendicularity:

4+4\lambda=0

Solving for

\lambda

\;4\lambda=-4

\;\lambda=-1

Thus, the correct value of

\lambda

is -1 .

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