Let $\overrightarrow{\mathrm{c}}$ be the projection vector of $\overrightarrow{\mathrm{b}}=\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{k}}, \lambda>0$, on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$, then the area of the parallelogram formed by the vectors $\vec{b}$ and $\overrightarrow{\mathrm{c}}$ is ______.

Vector Algebra — JEE Main · 2025 · 22 Jan 2025 · Shift 1 Question with Solution

Let

c

be the projection vector of

b = λ \hat{i} + 4 \hat{k}, λ > 0

, on the vector

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

. If

∣ a + c ∣ = 7

, then the area of the parallelogram formed by the vectors

b

and

c

is ______.

Answer: 16

The projection vector of

b = λ \hat{i} + 4 \hat{k}

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

c = (\frac{b \cdot a}{∣ a ∣ ^{2}}) a .

∣ a ∣^{2} = 1 + 4 + 4 = 9

, and

b \cdot a = λ + 8

, so

c = \frac{λ + 8}{9} (\hat{i} + 2 \hat{j} + 2 \hat{k}) .

a + c = a (1 + \frac{λ + 8}{9}) = a \cdot \frac{λ + 17}{9}

, and since

∣ a ∣ = 3

∣ a + c ∣ = 3 \cdot \frac{λ + 17}{9} = \frac{λ + 17}{3} = 7 ⟹ λ = 4.

b = 4 \hat{i} + 4 \hat{k}

and

c = \frac{12}{9} (\hat{i} + 2 \hat{j} + 2 \hat{k}) = \frac{4}{3} \hat{i} + \frac{8}{3} \hat{j} + \frac{8}{3} \hat{k} .

Area of the parallelogram formed by

b, c

∣ b \times c ∣

b \times c = \hat{i} 4 \frac{4}{3} \hat{j} 0 \frac{8}{3} \hat{k} 4 \frac{8}{3} = \hat{i} (0 \cdot \frac{8}{3} - 4 \cdot \frac{8}{3}) - \hat{j} (4 \cdot \frac{8}{3} - 4 \cdot \frac{4}{3}) + \hat{k} (4 \cdot \frac{8}{3} - 0)

= - \frac{32}{3} \hat{i} - \frac{16}{3} \hat{j} + \frac{32}{3} \hat{k} .

∣ b \times c ∣ = \frac{1}{3} 3 2^{2} + 1 6^{2} + 3 2^{2} = \frac{1}{3} 1024 + 256 + 1024 = \frac{1}{3} 2304 = \frac{48}{3} = 16.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer