Question

The point of intersection of the lines
\[ \vec r=2\vec b+t(6\vec c-\vec a) \] and
\[ \vec r=\vec a+s(\vec b-3\vec c) \] is

• A. \(\vec a+\vec b+\vec c\)
• B. \(\vec b-\vec c-6\vec a\)
• C. \(2\vec a-\vec b+\vec c\)
• D. \(\vec a+2\vec b-6\vec c\)

Hint:

For intersection of vector lines, expand both vector equations and compare coefficients of independent vectors.

Answer 1

The Correct Option is D

Step 1: Write both vector equations.
The given lines are
\[ \vec r=2\vec b+t(6\vec c-\vec a) \] and
\[ \vec r=\vec a+s(\vec b-3\vec c) \]
Expand both equations:
First line:
\[ \vec r=2\vec b+6t\vec c-t\vec a \]
Second line:
\[ \vec r=\vec a+s\vec b-3s\vec c \]

Step 2: Compare coefficients of \(\vec a,\vec b,\vec c\).
At the point of intersection, both expressions of \(\vec r\) are equal.
Thus, comparing coefficients:
Coefficient of \(\vec a\):
\[ -t=1 \] \[ t=-1 \]
Coefficient of \(\vec b\):
\[ 2=s \] \[ s=2 \]
Coefficient of \(\vec c\):
\[ 6t=-3s \]
Substituting \(t=-1\) and \(s=2\),
\[ 6(-1)=-3(2) \] \[ -6=-6 \] which is satisfied.

Step 3: Find the point of intersection.
Substitute \(t=-1\) into the first equation:
\[ \vec r = 2\vec b+(-1)(6\vec c-\vec a) \]
\[ = 2\vec b-6\vec c+\vec a \]
\[ = \vec a+2\vec b-6\vec c \]

Step 4: Final conclusion.
Hence, the point of intersection is
\[ \boxed{\vec a+2\vec b-6\vec c} \]