JEE Mathematics 2023 Question with Solution

Given: The vectors are

$$\lambda \mathbf{i}-\mathbf{j}+\mathbf{k},\quad \mathbf{i}+2\mathbf{j}+\mu\mathbf{k},\quad 3\mathbf{i}-4\mathbf{j}+5\mathbf{k}$$

with $\lambda-\mu=5$.

Find:

$$\sum_{(\lambda,\mu)\in S} 80(\lambda^2+\mu^2)$$

when the vectors are coplanar.

For coplanarity, the scalar triple product must be zero:

$$\begin{vmatrix} \lambda & -1 & 1 \\ 1 & 2 & \mu \\ 3 & -4 & 5 \end{vmatrix}=0$$

Evaluating the determinant,

$$\lambda \begin{vmatrix} 2 & \mu \\ -4 & 5 \end{vmatrix} +1 \begin{vmatrix} 1 & \mu \\ 3 & 5 \end{vmatrix} +1 \begin{vmatrix} 1 & 2 \\ 3 & -4 \end{vmatrix}=0$$ $$\lambda(10+4\mu)+(5-3\mu)+(-4-6)=0$$ $$\lambda(10+4\mu)-3\mu-5=0$$

Using the given relation,

$$\lambda=\mu+5$$

Substitute into the equation:

$$(\mu+5)(10+4\mu)-3\mu-5=0$$ $$4\mu^2+27\mu+45=0$$

Now solve:

$$\mu=\frac{-27\pm\sqrt{27^2-4\cdot4\cdot45}}{8}=\frac{-27\pm3}{8}$$

Hence,

$$\mu=-3,\quad -\frac{15}{4}$$

Corresponding values of $\lambda$ are

$$\lambda=2,\quad \frac{5}{4}$$

So the pairs are $(2,-3)$ and $\left(\frac{5}{4},-\frac{15}{4}\right)$.

For $(2,-3)$:

$$\lambda^2+\mu^2=4+9=13$$ $$80\times13=1040$$

For $\left(\frac{5}{4},-\frac{15}{4}\right)$:

$$\lambda^2+\mu^2=\frac{25}{16}+\frac{225}{16}=\frac{250}{16}$$ $$80\times\frac{250}{16}=1250$$

Therefore,

$$\sum_{(\lambda,\mu)\in S}80(\lambda^2+\mu^2)=1040+1250=2290$$

So, the correct option is C.

Given: The vectors are coplanar and satisfy $\lambda-\mu=5$.

Find: The required sum.

A quick way is to use $\lambda=\mu+5$ immediately in the scalar triple product condition. Since coplanarity means determinant zero,

$$\begin{vmatrix} \mu+5 & -1 & 1 \\ 1 & 2 & \mu \\ 3 & -4 & 5 \end{vmatrix}=0$$

This reduces the problem to one variable directly and gives

$$4\mu^2+27\mu+45=0$$

So,

$$\mu=-3,\quad -\frac{15}{4}$$

and hence

$$\lambda=2,\quad \frac{5}{4}$$

Now compute the two values of $80(\lambda^2+\mu^2)$:

$$80(2^2+(-3)^2)=1040$$ $$80\left(\left(\frac{5}{4}\right)^2+\left(-\frac{15}{4}\right)^2\right)=1250$$

Adding,

$$1040+1250=2290$$

Therefore, the correct option is C.

This works faster because the linear relation removes one variable before expanding everything.