Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let $$ \begin{array}{ll} \vec{a} = 3\hat{i} + \hat{j} - \hat{k}, & \\ \vec{b} = \hat{i} + b_2\hat{j} + b_3\hat{k}, & b_2, b_3 \in \mathbb{R}, \\ \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}, & c_1, c_2, c_3 \in \mathbb{R} \end{array} $$ be three vectors such that $b_2 b_3 > 0$, $\vec{a} \cdot \vec{b} = 0$ and $$ \left(\begin{array}{ccc} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ b_2 \\ b_3 \end{array}\right) = \left(\begin{array}{c} 3 - c_1 \\ 1 - c_2 \\ -1 - c_3 \end{array}\right). $$ Then, which of the following is/are TRUE?

(B) (B) $\vec{b} \cdot \vec{c} = 0$; (C) (C) $|\vec{b}| > \sqrt{10}$; (D) (D) $|\vec{c}| \leq \sqrt{11}$

(A) (A) $\vec{a} \cdot \vec{c} = 0$

Vector Algebra — JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 Question with Solution

Let

\hat{i}, \hat{j}

and

\hat{k}

be the unit vectors along the three positive coordinate axes. Let

a = 3 \hat{i} + \hat{j} - \hat{k}, b = \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}, b_{2}, b_{3} \in R, c_{1}, c_{2}, c_{3} \in R

be three vectors such that

b_{2} b_{3} > 0

a \cdot b = 0

and

0 c_{3} - c_{2} - c_{3} 0 c_{1} c_{2} - c_{1} 0 1 b_{2} b_{3} = 3 - c_{1} 1 - c_{2} - 1 - c_{3} .

Then, which of the following is/are TRUE?

Answer: (B), (C), (D)

The matrix equation gives:

b_{2} c_{3} - b_{3} c_{2} = c_{1} - 3

c_{3} - b_{3} c_{1} = 1 - c_{2}

c_{2} - b_{2} c_{1} = - 1 - c_{3}

These can be written as:

b \times c = c - a

Taking dot product with

b

0 = b \cdot c - a \cdot b = b \cdot c

b ⊥ c

. (B) is true. ✓

Taking dot product with

c

0 = ∣ c ∣^{2} - a \cdot c

a \cdot c = ∣ c ∣^{2} \neq = 0

(A) is false.

Squaring

b \times c = c - a

∣ b ∣^{2} ∣ c ∣^{2} = ∣ c ∣^{2} + ∣ a ∣^{2} - 2 a \cdot c = ∣ c ∣^{2} + 11 - 2∣ c ∣^{2} = 11 - ∣ c ∣^{2}

∣ c ∣^{2} (∣ b ∣^{2} + 1) = 11

∣ c ∣^{2} = \frac{11}{∣ b ∣ ^{2} + 1} \leq 11

∣ c ∣ \leq 11

(D) is true. ✓

From

a \cdot b = 0

3 + b_{2} - b_{3} = 0

, so

b_{2} - b_{3} = - 3

b_{2}^{2} + b_{3}^{2} - 2 b_{2} b_{3} = 9

b_{2}^{2} + b_{3}^{2} = 9 + 2 b_{2} b_{3} > 9

(since

b_{2} b_{3} > 0

)

∣ b ∣^{2} = 1 + b_{2}^{2} + b_{3}^{2} > 10

∣ b ∣ > 10

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options