Question:
If \((1,5,35)\), \((7,5,5)\), \((1,\lambda,7)\) and \((2\lambda,1,2)\) are coplanar, then the sum of all possible values of \(\lambda\) is :
If \((1,5,35)\), \((7,5,5)\), \((1,\lambda,7)\) and \((2\lambda,1,2)\) are coplanar, then the sum of all possible values of \(\lambda\) is :
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Four points are coplanar when the scalar triple product of the three vectors from one point to the others vanishes.
Updated On: Nov 4, 2026
- \(-\frac{44}{5}\)
- \(-\frac{39}{5}\)
- \(\frac{44}{5}\)
- \(\frac{39}{5}\)
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The Correct Option is C
Solution and Explanation
Take \(A(1,5,35)\): \(\vec{AB}=(6,0,-30)\), \(\vec{AC}=(0,\lambda-5,-28)\), \(\vec{AD}=(2\lambda-1,-4,-33)\).
Coplanarity: \(\begin{vmatrix}6&0&-30\\0&\lambda-5&-28\\2\lambda-1&-4&-33\end{vmatrix}=0\).
\(=6\left[-33(\lambda-5)-112\right]+30(\lambda-5)(2\lambda-1)=6(53-33\lambda)+30(2\lambda^2-11\lambda+5)\)
\(=60\lambda^2-528\lambda+468=0\Rightarrow5\lambda^2-44\lambda+39=0\).
Sum of roots \(=\frac{44}{5}\) (roots \(\lambda=1,\frac{39}{5}\)).
Answer: (3) \(\frac{44}{5}\)
Coplanarity: \(\begin{vmatrix}6&0&-30\\0&\lambda-5&-28\\2\lambda-1&-4&-33\end{vmatrix}=0\).
\(=6\left[-33(\lambda-5)-112\right]+30(\lambda-5)(2\lambda-1)=6(53-33\lambda)+30(2\lambda^2-11\lambda+5)\)
\(=60\lambda^2-528\lambda+468=0\Rightarrow5\lambda^2-44\lambda+39=0\).
Sum of roots \(=\frac{44}{5}\) (roots \(\lambda=1,\frac{39}{5}\)).
Answer: (3) \(\frac{44}{5}\)
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