Find $\lambda$ for which the points A(3, 2, 1), B(4, $\lambda$, 5), C(4, 2, -2) and D(6, 5, -1) are coplanar.
✓
Answer
The points A, B, C and D will be coplanar if will be coplanar of any one of the following traces of vectors are coplanar:
$\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}};\vec{\text{AB}},\vec{\text{BC}},\vec{\text{CD}};\vec{\text{BC}},\vec{\text{BA}},\vec{\text{BD}},$ etc.
It is given that $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar.
Thus, their scaler triple product $\big[\vec{\text{AB }}\vec{\text{AC }}\vec{\text{AD}}\big]$ is equal to zero.
Now,
Direction ratios of the $\vec{\text{PQ}}$ =(Direction ratios of vector Q) - (Direction ratios of the vector P)
Direction ratios of vector $\vec{\text{AB}}=(4-3),(\lambda-2),(5-1),\text{i. e. 1},\lambda, -2, 4$
Direction ratios of vector $\vec{\text{AC}}=(4-3),(2-2),(-2 -1),\text{i. e. } 3, 3, -2$
Direction ratios of vector $\vec{\text{AD}}=(6-3),(5-2),(-1-1),\text{i. e}. 3, 3, -2$
$\therefore\big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\big]=\begin{vmatrix}1&\lambda-2&4\\1&0&-3\\3&3&-2\end{vmatrix}$
$=1[0-(-9)]-(\lambda-2)[-2-(-9)]+4(3-0)=0$
$\Rightarrow7\lambda=35$
$\Rightarrow\lambda=5$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.