Q70 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics
If $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}},-12 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}},-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and $\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are the position vectors of four coplanar points, then $\lambda=$
- A-2
- Correct answerB6
- C3
- D-6
Solution
If four points are collinear, then
$\begin{aligned}<br />\left|\begin{array}{lll}<br />x_2-x_1 & y_2-y_1 & z_2-z_1 \\<br />x_3-x_1 & y_3-y_1 & z_3-z_1 \\<br />x_4-x_1 & y_4-y_1 & z_4-z_1<br />\end{array}\right| & =0 \\<br />\left|\begin{array}{ccc}<br />-14 & 0 & -6 \\<br />-3 & 3 & -7 \\<br />\lambda-2 & 3 & -4<br />\end{array}\right| & =0<br />\end{aligned}$
So, $-14(-12+21)-0(-)-6(-9-3 \lambda+16)=0$
$\begin{aligned}<br />-126-6(-3-3 \lambda) & =0 \\<br />-126+18+18 \lambda & =0 \\<br />18 \lambda & =+108 \\<br />\lambda & =6<br />\end{aligned}$
$\begin{aligned}<br />\left|\begin{array}{lll}<br />x_2-x_1 & y_2-y_1 & z_2-z_1 \\<br />x_3-x_1 & y_3-y_1 & z_3-z_1 \\<br />x_4-x_1 & y_4-y_1 & z_4-z_1<br />\end{array}\right| & =0 \\<br />\left|\begin{array}{ccc}<br />-14 & 0 & -6 \\<br />-3 & 3 & -7 \\<br />\lambda-2 & 3 & -4<br />\end{array}\right| & =0<br />\end{aligned}$
So, $-14(-12+21)-0(-)-6(-9-3 \lambda+16)=0$
$\begin{aligned}<br />-126-6(-3-3 \lambda) & =0 \\<br />-126+18+18 \lambda & =0 \\<br />18 \lambda & =+108 \\<br />\lambda & =6<br />\end{aligned}$
Hence, the correct answer is option: 'B'.