Let \mathop a\limits^ \to = 3\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + x\mathop k\limits^ \wedge and \mathop b\limits^ \to = \mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge , for some real x. Then | \mathop a\limits^ \to \times \mathop b\limits^ \to | = r is possible if :

3\sqrt 3 \over 2 \lt r \lt 5\sqrt 3 \over 2

\sqrt 3 \over 2 \lt r \le 3\sqrt 3 \over 2

Vector Algebra PYQ — JEE Main 2019 · 8 Apr · Shift 2 · Q25 with Solution

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Correct answer: C

$\vec a = 3\hat i + 2\hat j + x\hat k, \qquad \vec b = \hat i - \hat j + \hat k$

We need the possible values of

$r = |\vec a \times \vec b|.$

\begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 2 & x \\ 1 & -1 & 1 \end{vmatrix}$$ Expanding: $$\vec a \times \vec b = \hat i(2\cdot 1 - x(-1)) - \hat j(3\cdot 1 - x\cdot 1) + \hat k(3(-1)-2\cdot 1)$$ $$= \hat i(2+x) - \hat j(3-x) + \hat k(-5)$$ So, $$\vec a \times \vec b = (x+2)\hat i + (x-3)\hat j - 5\hat k.$$ --- 3. **Magnitude of the cross product** $$r = |\vec a \times \vec b| = \sqrt{(x+2)^2 + (x-3)^2 + (-5)^2}$$ Thus, $$r^2 = (x+2)^2 + (x-3)^2 + 25$$ $$= x^2+4x+4 + x^2-6x+9 + 25$$ $$= 2x^2 - 2x + 38.$$ --- 4. **Find the minimum possible value of** $r$ Since this is a quadratic in $x$, complete the square: $$r^2 = 2x^2 - 2x + 38 = 2\left(x^2-x\right)+38$$ $$= 2\left[\left(x-\frac12\right)^2 - \frac14\right] + 38$$ $$= 2\left(x-\frac12\right)^2 - \frac12 + 38$$ $$= 2\left(x-\frac12\right)^2 + \frac{75}{2}.$$ Therefore, $$r^2 \ge \frac{75}{2}$$ and hence $$r \ge \sqrt{\frac{75}{2}} = 5\sqrt{\frac32}.$$ Also, equality is attained when $$x = \frac12.$$ So the possible values are $$r \ge 5\sqrt{\frac32}.$$ --- 5. **Check options** - **A:** $0<r<\sqrt{\frac32}$ → not possible - **B:** $3\sqrt{\frac32}<r<5\sqrt{\frac32}$ → not possible - **C:** $r\ge 5\sqrt{\frac32}$ → possible - **D:** $\sqrt{\frac32}<r\le 3\sqrt{\frac32}$ → not possible Thus the correct option is: $$\boxed{\text{C}}$$

View written solutionFree

Correct answer: C

$\vec a = 3\hat i + 2\hat j + x\hat k, \qquad \vec b = \hat i - \hat j + \hat k$

We need the possible values of

$r = |\vec a \times \vec b|.$

\begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 2 & x \\ 1 & -1 & 1 \end{vmatrix}$$ Expanding: $$\vec a \times \vec b = \hat i(2\cdot 1 - x(-1)) - \hat j(3\cdot 1 - x\cdot 1) + \hat k(3(-1)-2\cdot 1)$$ $$= \hat i(2+x) - \hat j(3-x) + \hat k(-5)$$ So, $$\vec a \times \vec b = (x+2)\hat i + (x-3)\hat j - 5\hat k.$$ --- 3. **Magnitude of the cross product** $$r = |\vec a \times \vec b| = \sqrt{(x+2)^2 + (x-3)^2 + (-5)^2}$$ Thus, $$r^2 = (x+2)^2 + (x-3)^2 + 25$$ $$= x^2+4x+4 + x^2-6x+9 + 25$$ $$= 2x^2 - 2x + 38.$$ --- 4. **Find the minimum possible value of** $r$ Since this is a quadratic in $x$, complete the square: $$r^2 = 2x^2 - 2x + 38 = 2\left(x^2-x\right)+38$$ $$= 2\left[\left(x-\frac12\right)^2 - \frac14\right] + 38$$ $$= 2\left(x-\frac12\right)^2 - \frac12 + 38$$ $$= 2\left(x-\frac12\right)^2 + \frac{75}{2}.$$ Therefore, $$r^2 \ge \frac{75}{2}$$ and hence $$r \ge \sqrt{\frac{75}{2}} = 5\sqrt{\frac32}.$$ Also, equality is attained when $$x = \frac12.$$ So the possible values are $$r \ge 5\sqrt{\frac32}.$$ --- 5. **Check options** - **A:** $0<r<\sqrt{\frac32}$ → not possible - **B:** $3\sqrt{\frac32}<r<5\sqrt{\frac32}$ → not possible - **C:** $r\ge 5\sqrt{\frac32}$ → possible - **D:** $\sqrt{\frac32}<r\le 3\sqrt{\frac32}$ → not possible Thus the correct option is: $$\boxed{\text{C}}$$