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Let a⃗=i^+2j^+3k^,b⃗=−i^+2j^+k^,c⃗=3i^+j^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = -\hat{i} + 2\hat{j} + \hat{k}, \vec{c} = 3\hat{i} + \hat{j} be three vectors. If (a⃗+λb⃗)(\vec{a} + \lambda\vec{b}) is perpendicular to c⃗\vec{c}, then the value of λ\lambda is

Solution

✅ Correct Option: 2

a⃗=i^+2j^+3k^,b⃗=−i^+2j^+k^,c⃗=3i^+j^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = -\hat{i} + 2\hat{j} + \hat{k}, \quad \vec{c} = 3\hat{i} + \hat{j}

Since (a⃗+λb⃗)⊥c⃗(\vec{a} + \lambda\vec{b}) \perp \vec{c}, their dot product equals zero:

(a⃗+λb⃗)⋅c⃗=0(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0


a⃗+λb⃗=(i^+2j^+3k^)+λ(−i^+2j^+k^)\vec{a} + \lambda\vec{b} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k})

=(1−λ)i^+(2+2λ)j^+(3+λ)k^= (1 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}


Taking the dot product with c⃗=3i^+j^+0k^\vec{c} = 3\hat{i} + \hat{j} + 0\hat{k}:

(a⃗+λb⃗)⋅c⃗=3(1−λ)+1(2+2λ)+0(3+λ)(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 3(1 - \lambda) + 1(2 + 2\lambda) + 0(3 + \lambda)

=3−3λ+2+2λ= 3 - 3\lambda + 2 + 2\lambda

=5−λ= 5 - \lambda


5−λ=05 - \lambda = 0

λ=5\lambda = 5

Therefore, λ=5\boxed{\lambda = 5}

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