JEE Mathematics 2023 Question with Solution

Given: A($2, 6, 2$), B($-4, 0, \lambda$), C($2, 3, -1$), and D($4, 5, 0$) are vertices of a quadrilateral, with $|\lambda| \leq 5$ and area $18$ square units.

Find: The value of $5 - 6\lambda$.

For a quadrilateral in space with diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$,

$$\text{Area} = \frac{1}{2}\left|\overrightarrow{BD} \times \overrightarrow{AC}\right|$$

So,

$$\left|\overrightarrow{BD} \times \overrightarrow{AC}\right| = 36$$

Now,

$$\overrightarrow{AC} = (2-2, 3-6, -1-2) = (0, -3, -3)$$ $$\overrightarrow{BD} = (4-(-4), 5-0, 0-\lambda) = (8, 5, -\lambda)$$

Compute the cross product:

$$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{vmatrix}$$ $$= (3\lambda + 15)\hat{i} - (-24)\hat{j} + (-24)\hat{k}$$ $$= (3\lambda + 15)\hat{i} + 24\hat{j} - 24\hat{k}$$

Its magnitude is

$$\sqrt{(3\lambda + 15)^2 + 24^2 + 24^2} = 36$$

Squaring and simplifying gives

$$\lambda^2 + 10\lambda + 9 = 0$$ $$\lambda = -1, -9$$

Since $|\lambda| \leq 5$, we take

$$\lambda = -1$$

Therefore,

$$5 - 6\lambda = 5 - 6(-1) = 11$$

Therefore, the required value is $11$.

Given: The vertices are A($2, 6, 2$), B($-4, 0, \lambda$), C($2, 3, -1$), D($4, 5, 0$).

Find: $5 - 6\lambda$ when the area of quadrilateral $ABCD$ is $18$.

Use

$$\text{Area} = \frac{1}{2} \left| \overrightarrow{BD} \times \overrightarrow{AC} \right|$$

Thus,

$$\frac{1}{2} \left| \overrightarrow{BD} \times \overrightarrow{AC} \right| = 18$$

which implies

$$\left| \overrightarrow{BD} \times \overrightarrow{AC} \right| = 36$$

Now,

$$\overrightarrow{AC} = (0, -3, -3), \qquad \overrightarrow{BD} = (8, 5, -\lambda)$$

Then

$$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{vmatrix}$$ $$= (3\lambda + 15) \hat{i} + 24\hat{j} - 24\hat{k}$$

So,

$$\sqrt{(3\lambda + 15)^2 + 24^2 + (-24)^2} = 36$$

This gives

$$(3\lambda + 15)^2 + 24^2 + 24^2 = 36^2$$ $$(3\lambda + 15)^2 + 1152 = 1296$$ $$(3\lambda + 15)^2 = 144$$ $$3\lambda + 15 = \pm 12$$

Hence,

$$\lambda = -1 \text{ or } -9$$

Using $|\lambda| \leq 5$, we get $\lambda = -1$. Finally,

$$5 - 6\lambda = 5 + 6 = 11$$

Therefore, the required value is $11$.