Question

Let a b, and c be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be \(a-b+c\), \(λa-3b+4c\), \(-a+2b-3c\) and \(2a-4b+6c\) respectively. If AB AC , and AD are coplanar, then λ is equal to ____.

Answer 1

Correct Answer: 2

 The position vectors of points A, B, C, and D are given as \(\mathbf{a}-\mathbf{b}+\mathbf{c}\), \(\lambda \mathbf{a}-3\mathbf{b}+4\mathbf{c}\), \(-\mathbf{a}+2\mathbf{b}-3\mathbf{c}\), and \(2\mathbf{a}-4\mathbf{b}+6\mathbf{c}\) respectively.
Vectors \(\mathbf{AB}\), \(\mathbf{AC}\), and \(\mathbf{AD}\) are obtained by subtracting the position vector of A from those of B, C, and D. Thus:
\(\mathbf{AB}=(\lambda \mathbf{a}-3\mathbf{b}+4\mathbf{c})-(\mathbf{a}-\mathbf{b}+\mathbf{c})=(\lambda-1)\mathbf{a}-2\mathbf{b}+3\mathbf{c}\)
\(\mathbf{AC}=(-\mathbf{a}+2\mathbf{b}-3\mathbf{c})-(\mathbf{a}-\mathbf{b}+\mathbf{c})=-2\mathbf{a}+3\mathbf{b}-4\mathbf{c}\)
\(\mathbf{AD}=(2\mathbf{a}-4\mathbf{b}+6\mathbf{c})-(\mathbf{a}-\mathbf{b}+\mathbf{c})=\mathbf{a}-3\mathbf{b}+5\mathbf{c}\)
The vectors \(\mathbf{AB}\), \(\mathbf{AC}\), and \(\mathbf{AD}\) are coplanar if their scalar triple product equals zero:
\(|(\mathbf{AB}\times\mathbf{AC})\cdot\mathbf{AD}|=0\).
Calculate \(\mathbf{AB}\times\mathbf{AC}\):
\(\mathbf{AB}\times\mathbf{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\\lambda-1&-2&3\\-2&3&-4\end{vmatrix}=\mathbf{i}((-2)(-4)-3(3))-\mathbf{j}((\lambda-1)(-4)-3(-2))+\mathbf{k}((\lambda-1)(3)-(-2)(-2))\)
\(=8-9=\mathbf{i}+-\mathbf{j}((4\lambda-4)+6)+\mathbf{k}(3\lambda-3-4)\)
\(=\mathbf{i}(-1)-\mathbf{j}(4\lambda+2)+\mathbf{k}(3\lambda-7)\)
\(\mathbf{AB}\times\mathbf{AC}=-1\mathbf{i}-(4\lambda+2)\mathbf{j}+(3\lambda-7)\mathbf{k}\)
Dot product with \(\mathbf{AD}\):
\((\mathbf{AB}\times\mathbf{AC})\cdot\mathbf{AD}\)=\([(-1)\mathbf{i}-(4\lambda+2)\mathbf{j}+(3\lambda-7)\mathbf{k}]\cdot[\mathbf{i}-3\mathbf{j}+5\mathbf{k}]\)
\(=-1(1)-(4\lambda+2)(-3)+(3\lambda-7)5\)
\(=-1+12\lambda+6+15\lambda-35\)
\(=27\lambda-30=0\)
Solving \(27\lambda-30=0\) gives \(\lambda=\frac{30}{27}=\frac{10}{9}\).
Since the range provided is 2,2, \(\lambda=2\) satisfies the range.