Q74 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vector and the points $\lambda \vec{a}+3 \vec{b}-\vec{c}, \vec{a}-\lambda \vec{b}+3 \vec{c}, 3 \vec{a}+4 \vec{b}-\lambda \vec{c}, \vec{a}-6 \vec{b}+6 \vec{c}$ are coplanar, then one of the values of $\lambda$ is

  • A
    7
  • B
    5
  • Correct answer
    C
    2
  • D
    1

Solution

Vectors lying on the plane are $(\lambda-1) \vec{a}+9 \vec{b}+7 \vec{c} ;(6-\lambda) \vec{b}-3 \vec{c} ; 2 \vec{a}+10 \vec{b}-(\lambda+6) \vec{c}$
These 3 vectors are coplanar. $\Rightarrow\left|\begin{array}{ccc} \lambda-1 & 9 & -7 \\ 0 & 6-\lambda & -3 \\ 2 & 10 & -\lambda-6 \end{array}\right|=0$
Only $\lambda=2$ satisfies this equation.

Hence, the correct answer is option: 'C'.

Relevant Questions