If the area of a parallelogram, whose diagonals are $\hat{i} - \hat{j} + 2\hat{k}$ and $2\hat{i} + 3\hat{j} + \alpha \hat{k}$ is $\dfrac{\sqrt{93}}{2}$ sq. units, then find $\alpha$.

Question

The surface area of a spherical balloon is increasing at the rate of $ 2 \, \text{cm}^2/\text{sec} $. Then the rate of increase in the volume of the balloon, when the radius of the balloon is $ 6 \, \text{cm} $, is:

Answer 1

To solve the given question, we need to find the value of $\alpha$ using the information that the area of the parallelogram formed by two vectors representing its diagonals is $\frac{\sqrt{93}}{2}$ square units.

The area $A$ of a parallelogram formed by two diagonals $\mathbf{d_1}$ and $\mathbf{d_2}$ can be found using the formula:

$A = \frac{1}{2} \cdot |\mathbf{d_1} \times \mathbf{d_2}|$

Given:

$\mathbf{d_1} = \hat{i} - \hat{j} + 2\hat{k}$
$\mathbf{d_2} = 2\hat{i} + 3\hat{j} + \alpha \hat{k}$

The cross product $\mathbf{d_1} \times \mathbf{d_2}$ is calculated as:

$ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 3 & \alpha \end{vmatrix} $

This determinant evaluates to:

$\mathbf{d_1} \times \mathbf{d_2} = \hat{i}( (-1)\alpha - (3 \cdot 2) ) - \hat{j}( (1)\alpha - (2 \cdot 2) ) + \hat{k}( (1 \cdot 3) - ( (-1) \cdot 2) )$

Simplifying, we get:

$\mathbf{d_1} \times \mathbf{d_2} = \hat{i}( -\alpha - 6 ) - \hat{j}( \alpha - 4 ) + \hat{k}( 3 + 2 )$

$\mathbf{d_1} \times \mathbf{d_2} = \hat{i}( -\alpha - 6 ) - \hat{j}( \alpha - 4 ) + \hat{k}( 5 )$

The magnitude is calculated as:

$|\mathbf{d_1} \times \mathbf{d_2}| = \sqrt{(-\alpha - 6)^2 + (-\alpha + 4)^2 + 5^2}$

Substituting the given area:

$A = \frac{1}{2} \sqrt{(-\alpha - 6)^2 + (-\alpha + 4)^2 + 5^2} = \frac{\sqrt{93}}{2}$

Squaring both sides:

$(-\alpha - 6)^2 + (-\alpha + 4)^2 + 5^2 = 93$

Expanding the squares:

$(\alpha^2 + 12\alpha + 36) + (\alpha^2 - 8\alpha + 16) + 25 = 93$

Combining like terms:

$2\alpha^2 + 4\alpha + 77 = 93$

Solving for $\alpha$:

$2\alpha^2 + 4\alpha - 16 = 0$

Divide by 2:

$\alpha^2 + 2\alpha - 8 = 0$

Solving the quadratic equation:

$\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values $a = 1, b = 2, c = -8$:

$\alpha = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}$

$\alpha = \frac{-2 \pm \sqrt{4 + 32}}{2}$

$\alpha = \frac{-2 \pm \sqrt{36}}{2}$

$\alpha = \frac{-2 \pm 6}{2}$

Thus, the roots are $\alpha = \frac{-2 + 6}{2} = 2$ and $\alpha = \frac{-2 - 6}{2} = -4$.

Therefore, the correct answer is: -4, -2

If the area of a parallelogram, whose diagonals are $\hat{i} - \hat{j} + 2\hat{k}$ and $2\hat{i} + 3\hat{j} + \alpha \hat{k}$ is $\dfrac{\sqrt{93}}{2}$ sq. units, then find $\alpha$.

Show Hint

The Correct Option is B

Solution and Explanation

Top Questions on Integral Calculus

Questions Asked in MHT CET exam