Let \overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k , \overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k and \overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k be three vectors such that, | \overrightarrow b \times \overrightarrow c | = 5\sqrt 3 and \overrightarrow a is perpendicular to \overrightarrow b . Then the greatest amongst the values of | \overrightarrow a |^2 is \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q34 with Solution

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Correct answer: 90

Given vectors

$\vec a = \hat i + 5\hat j + \alpha \hat k = (1,5,\alpha)$ $\vec b = \hat i + 3\hat j + \beta \hat k = (1,3,\beta)$ $\vec c = -\hat i + 2\hat j - 3\hat k = (-1,2,-3)$

We are given:

$|\vec b \times \vec c| = 5\sqrt 3$
$\vec a \perp \vec b$

We need the greatest value of $|\vec a|^2$ .

Use perpendicularity condition

Since $\vec a \perp \vec b$ ,

$\vec a \cdot \vec b = 0$

So,

$1\cdot 1 + 5\cdot 3 + \alpha\beta = 0$ $1 + 15 + \alpha\beta = 0$ $\alpha\beta = -16$

Thus,

$\alpha = \frac{-16}{\beta}$

Use the cross product magnitude condition

Compute $\vec b \times \vec c$ :

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 3 & \beta\\ -1 & 2 & -3 \end{vmatrix}$$ Expanding, $$\vec b \times \vec c = \hat i(3(-3)-\beta\cdot 2) - \hat j(1(-3)-\beta(-1)) + \hat k(1\cdot 2 - 3(-1))$$ $$= \hat i(-9-2\beta) - \hat j(-3+\beta) + \hat k(2+3)$$ $$= (-9-2\beta)\hat i + (3-\beta)\hat j + 5\hat k$$ Now, $$|\vec b\times \vec c|^2 = (-9-2\beta)^2 + (3-\beta)^2 + 5^2$$ Given $$|\vec b\times \vec c| = 5\sqrt 3$$, $$|\vec b\times \vec c|^2 = 75$$ So, $$(-9-2\beta)^2 + (3-\beta)^2 + 25 = 75$$ $$(-9-2\beta)^2 + (3-\beta)^2 = 50$$ Expand: $$(81 + 36\beta + 4\beta^2) + (9 - 6\beta + \beta^2) = 50$$ $$5\beta^2 + 30\beta + 90 = 50$$ $$5\beta^2 + 30\beta + 40 = 0$$ $$\beta^2 + 6\beta + 8 = 0$$ $$(\beta+2)(\beta+4)=0$$ Hence, $$\beta = -2 \quad \text{or} \quad \beta = -4$$ --- 4. **Find corresponding values of** $$\alpha$$ Since $$\alpha\beta = -16$$: - If $$\beta=-2$$, $$\alpha = \frac{-16}{-2} = 8$$ - If $$\beta=-4$$, $$\alpha = \frac{-16}{-4} = 4$$ --- 5. **Compute** $$|\vec a|^2$$ $$|\vec a|^2 = 1^2 + 5^2 + \alpha^2 = 26 + \alpha^2$$ - For $$\alpha=8$$: $$|\vec a|^2 = 26 + 64 = 90$$ - For $$\alpha=4$$: $$|\vec a|^2 = 26 + 16 = 42$$ Therefore, the greatest value is $$\boxed{90}$$ --- 6. **Comparison with stored answer** Stored correct answer: $$90$$ Our derived answer matches it.

More from Vector Algebra

View written solutionFree

Correct answer: 90

Given vectors

$\vec a = \hat i + 5\hat j + \alpha \hat k = (1,5,\alpha)$ $\vec b = \hat i + 3\hat j + \beta \hat k = (1,3,\beta)$ $\vec c = -\hat i + 2\hat j - 3\hat k = (-1,2,-3)$

We are given:

$|\vec b \times \vec c| = 5\sqrt 3$
$\vec a \perp \vec b$

We need the greatest value of $|\vec a|^2$ .

Use perpendicularity condition

Since $\vec a \perp \vec b$ ,

$\vec a \cdot \vec b = 0$

So,

$1\cdot 1 + 5\cdot 3 + \alpha\beta = 0$ $1 + 15 + \alpha\beta = 0$ $\alpha\beta = -16$

Thus,

$\alpha = \frac{-16}{\beta}$

Use the cross product magnitude condition

Compute $\vec b \times \vec c$ :

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 3 & \beta\\ -1 & 2 & -3 \end{vmatrix}$$ Expanding, $$\vec b \times \vec c = \hat i(3(-3)-\beta\cdot 2) - \hat j(1(-3)-\beta(-1)) + \hat k(1\cdot 2 - 3(-1))$$ $$= \hat i(-9-2\beta) - \hat j(-3+\beta) + \hat k(2+3)$$ $$= (-9-2\beta)\hat i + (3-\beta)\hat j + 5\hat k$$ Now, $$|\vec b\times \vec c|^2 = (-9-2\beta)^2 + (3-\beta)^2 + 5^2$$ Given $$|\vec b\times \vec c| = 5\sqrt 3$$, $$|\vec b\times \vec c|^2 = 75$$ So, $$(-9-2\beta)^2 + (3-\beta)^2 + 25 = 75$$ $$(-9-2\beta)^2 + (3-\beta)^2 = 50$$ Expand: $$(81 + 36\beta + 4\beta^2) + (9 - 6\beta + \beta^2) = 50$$ $$5\beta^2 + 30\beta + 90 = 50$$ $$5\beta^2 + 30\beta + 40 = 0$$ $$\beta^2 + 6\beta + 8 = 0$$ $$(\beta+2)(\beta+4)=0$$ Hence, $$\beta = -2 \quad \text{or} \quad \beta = -4$$ --- 4. **Find corresponding values of** $$\alpha$$ Since $$\alpha\beta = -16$$: - If $$\beta=-2$$, $$\alpha = \frac{-16}{-2} = 8$$ - If $$\beta=-4$$, $$\alpha = \frac{-16}{-4} = 4$$ --- 5. **Compute** $$|\vec a|^2$$ $$|\vec a|^2 = 1^2 + 5^2 + \alpha^2 = 26 + \alpha^2$$ - For $$\alpha=8$$: $$|\vec a|^2 = 26 + 64 = 90$$ - For $$\alpha=4$$: $$|\vec a|^2 = 26 + 16 = 42$$ Therefore, the greatest value is $$\boxed{90}$$ --- 6. **Comparison with stored answer** Stored correct answer: $$90$$ Our derived answer matches it.

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