NIMCET 2016 — Mathematics PYQ

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Question

NIMCET 2016 — Mathematics PYQ

NIMCET | Mathematics | 2016

If $a, b$ are vectors such that $∣ a + b ∣ = 29$ and $a \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times b$ then possible value of $(a + b) . (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k})$ is

Choose the correct answer:

Explanation

Concept:
- The cross product of vector to itself = 0
- The cross product of collinear vectors = 0
- The dot product of collinear vectors = Product of their Magnitudes
- $a \times b = - b \times a$
- For dot product $(P + Q) \cdot R = (P \cdot R) + (Q \cdot R)$
- For cross product $(P + Q) \times R = (P \times R) + (Q \times R)$
- The unit vector in the direction of a $P = \hat{P} = \frac{P}{∣ P ∣}$
- A vector $X$ in direction of $P = (Magnitude of X) \times \hat{P}$

**Calculation:**
Given:
$a \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times b$

$\Rightarrow a \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times b = 0$

$\Rightarrow a \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + b \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = 0$

$\Rightarrow (a + b) \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = 0$

It means the $a + b$ and $(2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ are collinear vectors.

$∴ a + b = a + b \times \frac{2 i ^ + 3 j ^ + 4 k ^}{∣2 i ^ + 3 j ^ + 4 k ^ ∣}$

$\Rightarrow a + b = 29 \times \frac{2 i ^ + 3 j ^ + 4 k ^}{29}$

$\Rightarrow a + b = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$

$(a + b) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k})$

$\Rightarrow (a + b) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k}) = (2 \times (- 7) + 3 \times 2 + 4 \times 3)$

$\Rightarrow (a + b) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k}) = (- 14 + 6 + 12)$

$\Rightarrow (a + b) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k}) = 4$

Choose the correct answer:

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