Question

If \[ \vec r=\vec a+t\vec b \quad\text{and}\quad \vec r=\vec p+l\vec b \] are two lines, where \[ \vec a=\hat i+2\hat j,\qquad \vec b=\hat i+\hat j+\hat k,\qquad \vec p=2\hat i+\hat k. \] The position vector of a point \(A\) is \[ 5\hat i+\hat j-3\hat k. \] Let \(B,C\) be the feet of the perpendiculars drawn from \(A\) to the given two lines respectively, then \[ \overrightarrow{BA}+\overrightarrow{AC}= \]

• A. \(\hat i-2\hat j+3\hat k\)
• B. \(2\hat i-\hat j+3\hat k\)
• C. \(\hat i+\hat j+\hat k\)
• D. \(\hat i-2\hat j+\hat k\)

Hint:

The foot of the perpendicular from a point to a line is obtained by using the condition \[ \boxed{(\overrightarrow{PB})\cdot\vec d=0,} \] where \(\vec d\) is the direction vector of the line.

Answer 1

The Correct Option is D

Step 1: Find the foot of the perpendicular on the first line. The first line is \[ \vec r=(1,2,0)+t(1,1,1). \] Let \[ B=(1+t,\;2+t,\;t). \] Since \[ \overrightarrow{AB}\perp\vec b, \] we have \[ (B-A)\cdot(1,1,1)=0. \] Substituting \[ A=(5,1,-3), \] \[ (t-4)+(t+1)+(t+3)=0, \] \[ 3t=0, \] \[ t=0. \] Hence, \[ B=(1,2,0). \]

Step 2: Find the foot of the perpendicular on the second line. The second line is \[ \vec r=(2,0,1)+l(1,1,1). \] Let \[ C=(2+l,\;l,\;1+l). \] Again, \[ (C-A)\cdot(1,1,1)=0. \] Thus, \[ (l-3)+(l-1)+(l+4)=0, \] \[ 3l=0, \] \[ l=0. \] Hence, \[ C=(2,0,1). \]

Step 3: Find \(\overrightarrow{BA}+\overrightarrow{AC}\). Now, \[ \overrightarrow{BA}=A-B=(4,-1,-3), \] and \[ \overrightarrow{AC}=C-A=(-3,-1,4). \] Therefore, \[ \overrightarrow{BA}+\overrightarrow{AC} =(1,-2,1) =\hat i-2\hat j+\hat k. \] Hence, \[ \boxed{\hat i-2\hat j+\hat k}. \] Thus, \[ \boxed{(D)} \] is the correct answer.