Let three vectors \(\vec{a}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}\), \(\overrightarrow{\mathrm{b}}=5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}\) form a triangle such that \(\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}\) and the area of the triangle is \(5 \sqrt{6}\). if \(\alpha\) is a positive real number, then \(|\vec{c}|^{2}\) is :

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Vector Algebra

Let three vectors

a = α \hat{i} + 4 \hat{j} + 2 \hat{k}

b = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}, c = x \hat{i} + y \hat{j} + z \hat{k}

form a triangle such that

c = a - b

and the area of the triangle is

56

. if

α

is a positive real number, then

∣ c ∣^{2}

is :

Answer: (B)

Since

c = a - b

, the triangle has area

\frac{1}{2} ∣ a \times b ∣ = 56 \Rightarrow ∣ a \times b ∣ = 106

. With

a = (α, 4, 2), b = (5, 3, 4)

a \times b = (10, 10 - 4 α, 3 α - 20)

, so

(4 α - 10)^{2} + (3 α - 20)^{2} = 500 \Rightarrow 25 α^{2} - 200 α = 0 \Rightarrow α = 8

(positive). Then

c = a - b = (3, 1, - 2)

and

∣ c ∣^{2} = 9 + 1 + 4 = 14

— option (2).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option