NIMCET 2022 — Mathematics PYQ

The scalar triple product $[\vec{a}\vec{b}\vec{c}]$ is calculated as the determinant of a $3\times 3$ matrix formed by the components of vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Given:

$$\vec{a}=(\lambda ,1,-2)$$

$$\vec{b}=(1,\lambda ,-2)$$

$$\vec{c}=(1,1,1)$$

Set the determinant equal to $7$:

$$\left|\begin{array}{ccc}\lambda & amp;1& amp;-2\\ 1& amp;\lambda & amp;-2\\ 1& amp;1& amp;1\end{array}\right|=7$$

1. Expanding the Determinant

Expanding along the first row:

$$\lambda (\lambda (1)-(-2)(1))-1(1(1)-(-2)(1))+(-2)(1(1)-\lambda (1))=7$$

$$\lambda (\lambda +2)-1(1+2)-2(1-\lambda )=7$$

2. Simplifying the Equation

$${\lambda }^2+2\lambda -3-2+2\lambda =7$$

$${\lambda }^2+4\lambda -5=7$$

$${\lambda }^2+4\lambda -12=0$$

3. Solving for $\lambda$

Factor the quadratic equation:

$$(\lambda +6)(\lambda -2)=0$$

This gives us:

$$\lambda =-6\text{ or }\lambda =2$$

Correct Option: A) $2,-6$