Given that,
\(\vec a = 8 \vec b \) and \(\vec c = - 7\vec b\)
Let
\(b = b_1 \hat i + b_2 \hat j + b_3 \hat k\)
Then,
\(\vec a = 8(b_1 \hat i + b_2\hat j + b_3 \hat k)\)
And
\(\vec c = -7 (b_1\hat i + b_2 \hat j + b_3 \hat k)\)
Consider that angle between \(\vec a \) and \(\vec c\) is x.
Thus,
\(\cos x = \frac{a.c}{|a| |c|}\)
\(\cos x = \frac {-56 (b_1^2 + b_2^2 + b_3^2)}{7 \times 8\sqrt{b_1^2 + b_2^2 + b_3^2. } \sqrt{b_1^2 + b_2^2 + b_3^2}}\)
\(\cos x = −1\)
Therefore,
\(x = π\)
Hence angle between \(\vec a\) and \(\vec c\) is π.
