Question

The straight line given by the equation \[ \vec r=(4\hat{i}+5\hat{j}+\hat{k})+s(4\hat{i}+6\hat{j}+2\hat{k}) \] is coplanar with the straight line given below. Choose the correct option.

• A. \(\vec r=(\hat{i}-2\hat{j}+3\hat{k})+p(2\hat{i}+3\hat{j}-4\hat{k})\)
• B. \(\vec r=(3\hat{i}-4\hat{j}+3\hat{k})+q(-4\hat{i}+5\hat{j}-6\hat{k})\)
• C. \(\vec r=(2\hat{i}+5\hat{j}-4\hat{k})+r(\hat{i}+4\hat{j}-3\hat{k})\)
• D. \(\vec r=(-4\hat{i}+4\hat{j}+4\hat{k})+t(7\hat{i}+5\hat{j})\)

Hint:

Two lines are coplanar if \[ (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=0. \] This is the standard scalar triple product test for coplanarity.

Answer 1

The Correct Option is D

Step 1: Recall the condition for coplanarity of two lines.

If& nbsp;

\[ \vec r=\vec a_1+\lambda\vec b_1 \]

and

\[ \vec r=\vec a_2+\mu\vec b_2, \]

then the lines are coplanar if

\[ (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=0. \]

Step 2: Check option (D).

For the given line,

\[ \vec a_1=(4,5,1), \qquad \vec b_1=(4,6,2). \]

For option (D),

\[ \vec a_2=(-4,4,4), \qquad \vec b_2=(7,5,0). \]

Therefore,

\[ \vec a_2-\vec a_1=(-8,-1,3). \]

Also,

\[ \vec b_1\times\vec b_2 = \begin{vmatrix} \hat{i} & amp; \hat{j} & amp; \hat{k}\\ 4 & amp; 6 & amp; 2\\ 7 & amp; 5 & amp; 0 \end{vmatrix} = -10\hat{i}+14\hat{j}-22\hat{k}. \]

Hence,

\[ (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) = (-8)(-10)+(-1)(14)+3(-22) = 80-14-66 = 0. \]

Step 3: Conclude.

Since

\[ (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=0, \]

the two lines are coplanar.

Therefore,

\[ \boxed{\text{Option (D) is correct}.} \]