Let S be the set of all \((\lambda, \mu)\) for which the vectors \(\lambda \hat{i}-\hat{j}+\hat{k}, \hat{i}+2 \hat{j}+\mu \hat{k}\) and \(3 \hat{i}-4 \hat{j}+5 \hat{k}\), where \(\lambda-\mu=5\), are coplanar, then \(\sum_{(\lambda, \mu) \in \mathrm{S}} 80\left(\lambda^{2}+\mu^{2}\right)\) is equal to :

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Vector Algebra

Let S be the set of all

(λ, μ)

for which the vectors

λ \hat{i} - \hat{j} + \hat{k}, \hat{i} + 2 \hat{j} + μ \hat{k}

and

3 \hat{i} - 4 \hat{j} + 5 \hat{k}

, where

λ - μ = 5

, are coplanar, then

\sum_{(λ, μ) \in S} 80 (λ^{2} + μ^{2})

is equal to :

Answer: (C)

Three vectors are coplanar when their scalar triple product vanishes, so

λ 13 - 1 2 - 4 1 μ 5 = 0

. Expanding along the first row,

λ (10 + 4 μ) + (5 - 3 μ) + (- 4 - 6) = 0

, that is

4 λ μ + 10 λ - 3 μ - 5 = 0

. The constraint gives

λ = μ + 5

; substituting,

4 (μ + 5) μ + 10 (μ + 5) - 3 μ - 5 = 0

, which reduces to

4 μ^{2} + 27 μ + 45 = 0

, i.e.

(μ + 3) (4 μ + 15) = 0

. Hence

μ = - 3, λ = 2

μ = - \frac{15}{4}, λ = \frac{5}{4}

. Therefore

\sum 80 (λ^{2} + μ^{2}) = 80 [(4 + 9) + (\frac{25}{16} + \frac{225}{16})] = 80 (13 + \frac{250}{16}) = 1040 + 1250 = 2290

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option