Question

If the equation of the plane containing the line \[ \vec r=\hat i+2\hat j+\hat k+t(\hat i-\hat j+2\hat k) \] and parallel to the line \[ \vec r=-\hat i+2\hat j+s(-\hat i+2\hat j+\hat k) \] in Cartesian coordinates is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, \] then \(a+3b+c=\)

• A. \(2\)
• B. \(10\)
• C. \(-5\)
• D. \(12\)

Hint:

If a plane contains one line and is parallel to another line, use the direction vectors of both lines to obtain the normal vector through their cross product.

Answer 1

The Correct Option is A

Concept: A plane containing a line and parallel to another line contains two direction vectors. The normal vector of the plane is obtained from the cross product of these direction vectors.

Step 1: Identify a point and direction vectors. The given line is \[ \vec r=(1,2,1)+t(1,-1,2). \] Hence, a point on the plane is \[ P(1,2,1). \] The plane contains the direction vector \[ \vec d_1=(1,-1,2). \] The second line is parallel to the plane and has direction vector \[ \vec d_2=(-1,2,1). \] Thus, the plane contains both vectors \[ (1,-1,2) \quad\text{and}\quad (-1,2,1). \]

Step 2: Find the normal vector of the plane. \[ \vec n = \vec d_1\times \vec d_2. \] \[ = \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & 2\\ -1 & 2 & 1 \end{vmatrix}. \] \[ = \hat i(-1-4) -\hat j(1+2) +\hat k(2-1). \] \[ =(-5,-3,1). \] Thus, a normal vector is \[ (5,3,-1). \]

Step 3: Find the equation of the plane. Using point-normal form, \[ 5(x-1)+3(y-2)-(z-1)=0. \] \[ 5x+3y-z-10=0. \] \[ 5x+3y-z=10. \]

Step 4: Convert to intercept form. Dividing by \(10\), \[ \frac{x}{2}+\frac{y}{\frac{10}{3}}+\frac{z}{-10}=1. \] Hence, \[ a=2,\qquad b=\frac{10}{3},\qquad c=-10. \]

Step 5: Calculate \(a+3b+c\). \[ a+3b+c = 2+3\left(\frac{10}{3}\right)-10. \] \[ = 2+10-10. \] \[ =2. \] \[ \boxed{a+3b+c=2} \] \[ \boxed{\text{Answer = (A)}} \]