Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \quad \vec{b}=-\hat{i}-8 \hat{j}+2 \hat{k} \quad$ and $\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}+\mathrm{c}_{2} \hat{\mathrm{j}}+\mathrm{c}_{3} \hat{\mathrm{k}}$ be three vectors such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$. If the angle between the vector $\overrightarrow{\mathrm{c}}$ and the vector $3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ is $\theta$, then the greatest integer less than or equal to $\tan ^{2} \theta$ is :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Vector Algebra

Let

a = \hat{i} + \hat{j} + \hat{k}, b = - \hat{i} - 8 \hat{j} + 2 \hat{k}

and

c = 4 \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}

be three vectors such that

b \times a = c \times a

. If the angle between the vector

c

and the vector

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

θ

, then the greatest integer less than or equal to

tan^{2} θ

is :

Answer: 38

b \times a = c \times a ⟹ (b - c) \times a = 0

, so

b - c

is parallel to

a

b - c = λ a

for some scalar

λ

b - c = (- 1 - 4, - 8 - c_{2}, 2 - c_{3}) = (- 5, - 8 - c_{2}, 2 - c_{3}) = λ (1, 1, 1) .

Matching components:

- 5 = λ

, so

λ = - 5

- 8 - c_{2} = - 5 ⟹ c_{2} = - 3

2 - c_{3} = - 5 ⟹ c_{3} = 7

.

So

c = (4, - 3, 7)

cos θ = \frac{c \cdot ( 3 , 4 , 1 )}{∣ c ∣ ∣ ( 3 , 4 , 1 ) ∣} = \frac{12 - 12 + 7}{16 + 9 + 49 9 + 16 + 1} = \frac{7}{74 26} = \frac{7}{1924} .

tan^{2} θ = sec^{2} θ - 1 = \frac{1}{cos ^{2} θ} - 1 = \frac{1924}{49} - 1 = \frac{1875}{49} \approx 38.27.

The greatest integer less than or equal to

tan^{2} θ

38

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer