JEE Advanced · Mathematics · 30. Vector Algebra

Let $\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$. If for some real numbers $\alpha, \beta$, and $\gamma$, we have $15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} \times \vec{q})$ then the value of $\gamma$ is

Verified Solution

Answer & Solution

Correct Answer

(B) 2

Step-by-step Solution

Detailed explanation

$15 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=\alpha(2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})+\beta(\overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}})+\gamma(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$
taking dot with $(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$
$\left|\begin{array}{ccc}15 & 10 & 6 \\ 2 & 1 & 3 \\ 1 & -1 & 1\end{array}\right|=0+0+\gamma\left(\mathrm{p}^2 \mathrm{q}^2-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2\right) \quad\left[\because(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})^2+(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2=\mathrm{p}^2 \mathrm{q}^2\right]$
$\begin{aligned} & \Rightarrow 52=26 \gamma \\ & \therefore \gamma=2\end{aligned}$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let

f_{1} : R \to R, f_{2} (- \frac{π}{2}, \frac{π}{2}) \to R, f_{3} : (- 1, e^{\frac{π}{2}} - 2) \to R

and

f_{4} : ℝ \to ℝ

be functions defined by
(i)

f_{1} (x) = \sin (\sqrt{1 - e^{- x^{2}}})

(ii)

f_{2} (x) = \{\begin{cases} \frac{|\sin x|}{\tan^{- 1} (x)} & , x \neq 0 \\ 1 & , x = 0 \end{cases}

, where the inverse trigonometric function

\tan^{- 1} x

assumes values in

(- \frac{π}{2}, \frac{π}{2})

(iii)

f_{3} (x) = [\sin (\log_{e} (x + 2))]

, where, for

t \in R, [t]

denotes the greatest integer less than or equal to

t

,
(iv)

f_{4} (x) = \{\begin{matrix} x^{2} \sin (\frac{1}{x}) & , x \neq 0 \\ 0 & , x = 0 \end{matrix}

LIST-I	LIST-II
A. the function $f_{1}$ is	P. NOT continuous at $x = 0$
B. The function $f_{2}$ is	Q. continuous at $x = 0$ and NOT differentiable at $x = 0$
C. The function $f_{3}$ is	R. differentiable at $x = 0$ and its derivative is NOT continuous at $x = 0$
D. The function $f_{4}$ is	S. differentiable at $x = 0$ and its derivative is continuous at $x = 0$

The correct option is :

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