Referred to the cartesian axes with origin O, the equation of a line $\ell$ is $\displaystyle\frac{x-2}{1} = \frac{y+3}{-1} = \frac{z-5}{-1}$
The vector equation of line $\ell$ would be r= $2$i - $3$j + $5$k + $\lambda$ (i - j - k).
It is given that the point $P (0,-1,7)$ lies on $\ell$.
Find the size of the angle between $OP$ and the line $\ell$, giving your answer to the nearest $0.1$°.
To calculate the size of angle between two lines, I understand that I need to take the scalar product of two vectors, $\cos\theta = \displaystyle\frac{|b_1\cdot b_2|}{|b_1||b_2|}$
From the equation given, I know the direction of line $\ell$, which is b$_1$ = i - j - k. But I don’t know how to get the direction of line $OP$ in order to solve this question.
So, I took the direction of line $OP$ to be b$_2$ = -j + $7$k.
$\cos\theta=\displaystyle\frac{ \left( \begin {matrix} 1\\-1\\-1\\\end {matrix}\right)\cdot \left(\begin {matrix}0\\-1\\7\\\end {matrix}\right)}{(\sqrt{1+1+1})(\sqrt{0+1+49})} = \frac{1-7}{\sqrt3 \times \sqrt{50}}=\frac{-6}{(\sqrt3)(\sqrt{50})}$
$\theta=60.67°$
But, the answer given is $60.1°$. Is my method wrong?