NIMCET 2008 — Mathematics PYQ

Let the four points be $A,B,C,$ and $D$ with position vectors:

• $\overrightarrow{OA}=3\hat{i}-2\hat{j}+\lambda \hat{k}$

• $\overrightarrow{OB}=6\hat{i}+3\hat{j}+\hat{k}$

• $\overrightarrow{OC}=5\hat{i}+7\hat{j}+3\hat{k}$

• $\overrightarrow{OD}=2\hat{i}+2\hat{j}+6\hat{k}$

Step 2: Find the vectors lying in the plane

If four points are coplanar, then the three vectors formed by connecting one point to the other three must be linearly dependent. Let's find vectors $\overrightarrow{AB},\overrightarrow{AC},$ and $\overrightarrow{AD}$:

• $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(6-3)\hat{i}+(3-(-2))\hat{j}+(1-\lambda )\hat{k}=3\hat{i}+5\hat{j}+(1-\lambda )\hat{k}$

• $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(5-3)\hat{i}+(7-(-2))\hat{j}+(3-\lambda )\hat{k}=2\hat{i}+9\hat{j}+(3-\lambda )\hat{k}$

• $\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=(2-3)\hat{i}+(2-(-2))\hat{j}+(6-\lambda )\hat{k}=-1\hat{i}+4\hat{j}+(6-\lambda )\hat{k}$

Step 3: Condition for Coplanarity

For these vectors to be coplanar, their Scalar Triple Product must be zero:

This is equivalent to the determinant being zero:

Step 4: Solve the Determinant

Expanding along the first row:

Simplify inside the brackets:

Step 5: Simplify and solve for $\lambda$

Note that $-15\lambda$ and $+15\lambda$ cancel out:

Conclusion:

The value of $\lambda$ for which the points are coplanar is 4.

Correct Option: (b)