Question 155 of 379MediumMCQ
If the points with position vectors 3i^−4j^+2k^, i^+2j^−k^, −2i^−j^+3k^, 5i^−2αj^+4k^ are coplanar, α equals Solutions · 1
Official solution
Taking A=(3,−4,2) as base: AB=(−2,6,−3), AC=(−5,3,1), AD=(2,4−2α,2).
\[\begin{vmatrix}-2&6&-3\\-5&3&1\\2&4-2\alpha&2\end{vmatrix}=0\]
\[-2[6-(4-2\alpha)] - 6[-10-2] -3[-5(4-2\alpha)-6]=0\]
\[-2(2+2\alpha)+72-3(-26+10\alpha)=0 \implies -4-4\alpha+72+78-30\alpha=0 \implies 146=34\alpha\]
\[\alpha = \frac{73}{17}\] Answer the question to unlock