Step 1: Given,
\( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Rightarrow \vec{a} \cdot (\vec{b} - \vec{c}) = 0 \)
So, \( \vec{a} \) is perpendicular to \( (\vec{b} - \vec{c}) \).
Step 2: Given,
\( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \Rightarrow \vec{a} \times (\vec{b} - \vec{c}) = \vec{0} \)
So, \( \vec{a} \) is parallel to \( (\vec{b} - \vec{c}) \).
Step 3:
So we have:
- \( \vec{b} - \vec{c} \) is perpendicular to \( \vec{a} \)
- \( \vec{b} - \vec{c} \) is also parallel to \( \vec{a} \)
The only vector that can be both perpendicular and parallel to the same non-zero vector is the zero vector.
Therefore,
\( \vec{b} - \vec{c} = \vec{0} \Rightarrow \vec{b} = \vec{c} \)
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is: