Question

Let \[ A=(2,0,3),\qquad B=(0,1,4),\qquad C=(5,6,0) \] be three points. If \(L_1\) and \(L_2\) are the lines bisecting the angles between \(AB\) and \(AC\), then the direction ratios of the line perpendicular to both \(L_1\) and \(L_2\) are

• A. \((3,1,2)\)
• B. \((1,5,2)\)
• C. \((3,1,5)\)
• D. \((1,3,5)\)

Hint:

If two lines have direction vectors \[ \vec d_1,\qquad \vec d_2, \] then a line perpendicular to both has direction vector \[ \boxed{\vec d_1\times\vec d_2.} \]

Answer 1

The Correct Option is C

Step 1: Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). \[ \overrightarrow{AB} = (0-2,\;1-0,\;4-3) = (-2,1,1), \] \[ \overrightarrow{AC} = (5-2,\;6-0,\;0-3) = (3,6,-3). \] Their magnitudes are \[ |\overrightarrow{AB}|=\sqrt6, \qquad |\overrightarrow{AC}|=3\sqrt6. \] Hence the corresponding unit vectors are \[ \hat{u} = \left(-\frac2{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6}\right), \] \[ \hat{v} = \left(\frac1{\sqrt6},\frac2{\sqrt6},-\frac1{\sqrt6}\right). \]

Step 2: Find the direction vectors of the angle bisectors. The internal and external angle bisectors are along \[ \hat{u}+\hat{v} = \frac1{\sqrt6}(-1,3,0), \] and \[ \hat{u}-\hat{v} = \frac1{\sqrt6}(-3,-1,2). \] Thus, \[ L_1\parallel(-1,3,0), \qquad L_2\parallel(-3,-1,2). \]

Step 3: Find a vector perpendicular to both bisectors. A vector perpendicular to both is their cross product: \[ (-1,3,0)\times(-3,-1,2) = (6,2,10). \] Dividing by \(2\), \[ (6,2,10) = 2(3,1,5). \] Hence the required direction ratios are \[ \boxed{(3,1,5).} \] Therefore, the correct option is \(\boxed{(C)}\).