Source: Extra Practice

For what value of mm are the three vectors A⃗=2i^−j^+k^\vec{A} = 2\hat{i} - \hat{j} + \hat{k}, B⃗=i^+2j^−3k^\vec{B} = \hat{i} + 2\hat{j} - 3\hat{k} and C⃗=3i^+mj^+5k^\vec{C} = 3\hat{i} + m\hat{j} + 5\hat{k} coplanar?

Options

Option A is correct

−4-4

Option B

44

Option C

−2-2

Option D

22

Explanation

Three vectors are coplanar if their scalar triple product is zero: [A⃗  B⃗  C⃗]=0[\vec{A} \; \vec{B} \; \vec{C}] = 0. This can be evaluated using the determinant of their components: ∣2−1112−33m5∣=0\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & m & 5 \end{vmatrix} = 0. Expanding along the first row: 2(10−(−3m))−(−1)(5−(−9))+1(m−6)=0  ⟹  2(10+3m)+1(14)+m−6=0  ⟹  20+6m+14+m−6=0  ⟹  7m+28=0  ⟹  m=−42(10 - (-3m)) - (-1)(5 - (-9)) + 1(m - 6) = 0 \implies 2(10 + 3m) + 1(14) + m - 6 = 0 \implies 20 + 6m + 14 + m - 6 = 0 \implies 7m + 28 = 0 \implies m = -4. Thus, m=−4m = -4 is correct, and other values of mm will make the scalar triple product non-zero, meaning the vectors would not be coplanar.