Question

Let a line \(L\) passing through the point \((1,1,1)\) be perpendicular to both the vectors \(2\hat{i}+2\hat{j}+\hat{k}\) and \(\hat{i}+2\hat{j}+2\hat{k}\). If \((a,b,c)\) is the foot of perpendicular from the origin on the line \(L\), then the value of \(34(a+b+c)\) is:

• A. \(50\)
• B. \(80\)
• C. \(100\)
• D. \(120\)

Answer 1

The Correct Option is C

Concept: If a line is perpendicular to two vectors, then its direction vector is given by the cross product of those vectors. Also, the foot of the perpendicular from the origin to a line can be found using projection.
Step 1:Find the direction vector of the line.
The given vectors are \[ \vec{v}_1=(2,2,1), \quad \vec{v}_2=(1,2,2) \] Direction vector of the required line: \[ \vec{d}=\vec{v}_1 \times \vec{v}_2 \] \[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 2 & 1 \\ 1 & 2 & 2 \end{vmatrix} \] \[ =(2\cdot2-1\cdot2)\hat{i} -(2\cdot2-1\cdot1)\hat{j} +(2\cdot2-2\cdot1)\hat{k} \] \[ =(2,-3,2) \] Thus direction vector of line \(L\) is \[ (2,-3,2) \]
Step 2:Write equation of the line. Since the line passes through \((1,1,1)\), \[ \vec{r}=(1,1,1)+t(2,-3,2) \] \[ (x,y,z)=(1+2t,\;1-3t,\;1+2t) \]
Step 3:Use the perpendicular condition.} The foot of perpendicular from origin occurs when \[ \vec{r}\cdot\vec{d}=0 \] \[ (1+2t,1-3t,1+2t)\cdot(2,-3,2)=0 \] \[ 2(1+2t)-3(1-3t)+2(1+2t)=0 \] \[ 2+4t-3+9t+2+4t=0 \] \[ 1+17t=0 \] \[ t=-\frac{1}{17} \]
Step 4:Find the coordinates of the foot.} \[ a=1+\frac{-2}{17}=\frac{15}{17} \] \[ b=1+\frac{3}{17}=\frac{20}{17} \] \[ c=1+\frac{-2}{17}=\frac{15}{17} \] Thus \[ a+b+c=\frac{50}{17} \]
Step 5:Compute the required value. \[ 34(a+b+c)=34\left(\frac{50}{17}\right) \] \[ =2\times50 \] \[ =100 \]