Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k}$ and $\vec{d}=\vec{a} \times \vec{b}$. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-2 \overrightarrow{\mathrm{a}}|^{2}=8$ and the angle between $\overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}}$ is $\frac{\pi}{4}$, then $|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^{2}$ is equal to

Vector Algebra — JEE Main · 2025 · 28 Jan 2025 · Shift 1 Question with Solution

Let

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

and

d = a \times b

. If

c

is a vector such that

a \cdot c = ∣ c ∣, ∣ c - 2 a ∣^{2} = 8

and the angle between

d

and

c

\frac{π}{4}

, then

∣10 - 3 b \cdot c ∣ + ∣ d \times c ∣^{2}

is equal to

Answer: 6

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

d = a \times b = \hat{i} 12 \hat{j} 12 \hat{k} 11 = - \hat{i} + \hat{j}

, so

∣ d ∣^{2} = 2

**Finding

∣ c ∣

:** Expand

∣ c - 2 a ∣^{2} = 8

∣ c ∣^{2} + 4∣ a ∣^{2} - 4 (a \cdot c) = 8

Since

∣ a ∣^{2} = 3

and

a \cdot c = ∣ c ∣

∣ c ∣^{2} + 12 - 4∣ c ∣ = 8 \Rightarrow (∣ c ∣ - 2)^{2} = 0 \Rightarrow ∣ c ∣ = 2

**Finding

b \cdot c

:** Using

(a \times b) \times c = (a \cdot c) b - (b \cdot c) a

, with

a \cdot c = 2

d \times c = 2 b - (b \cdot c) a

Since the angle between

d

and

c

\frac{π}{4}

∣ d \times c ∣^{2} = ∣ d ∣^{2} ∣ c ∣^{2} sin^{2} \frac{π}{4} = 2 \cdot 4 \cdot \frac{1}{2} = 4

Also, expanding

∣2 b - (b \cdot c) a ∣^{2}

with

∣ b ∣^{2} = 9, a \cdot b = 5

4 = 4 (9) - 4 (b \cdot c) (5) + 3 (b \cdot c)^{2}

3 (b \cdot c)^{2} - 20 (b \cdot c) + 32 = 0 \Rightarrow b \cdot c = 4 or \frac{8}{3}

Either value gives the same result:

If

b \cdot c = 4

∣10 - 3 (4) ∣ + 4 = 2 + 4 = 6

b \cdot c = \frac{8}{3}

∣10 - 8∣ + 4 = 2 + 4 = 6

∣10 - 3 b \cdot c ∣ + ∣ d \times c ∣^{2} = 6

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Integer answer