Vector Algebra
Coplanarity of Vectors
Grade 12
Question:
<p>Let <b>a</b> = î + ĵ + k̂, <b>b</b> = î − ĵ + 2k̂ and <b>c</b> = xî + (x−2)ĵ − k̂. If the vector <b>c</b> lies in the plane of <b>a</b> and <b>b</b>, then x equals</p>
<p>0</p>
<p>1</p>
<p>−4</p>
<p>−2</p>
Step-by-Step Solution
Key Concept: A vector c lies in the plane of vectors a and b if and only if c is a linear combination of a and b, which means the scalar triple product a·(b×c) = 0.
Step 1: For vector c to lie in the plane of a and b, the vectors must be coplanar. This means: $\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$ Step 2: Expand along the first row: $1\begin{vmatrix} -1 & 2 \\ x-2 & -1 \end{vmatrix} - 1\begin{vmatrix} 1 & 2 \\ x & -1 \end{vmatrix} + 1\begin{vmatrix} 1 & -1 \\ x & x-2 \end{vmatrix} = 0$ Step 3: Calculate each 2×2 determinant: $1[(−1)(−1) − 2(x−2)] − 1[(1)(−1) − 2x] + 1[(1)(x−2) − (−1)x] = 0$ $1[1 − 2x + 4] − 1[−1 − 2x] + 1[x − 2 + x] = 0$ $(5 − 2x) − (−1 − 2x) + (2x − 2) = 0$ $5 − 2x + 1 + 2x + 2x − 2 = 0$ $2x + 4 = 0$ $x = -2$ ∴ Answer: D (x = -2)
Correct Answer: D