Let $Q$ be the point $(1,2,3)$, let $\overrightarrow{b} = \langle -1, 0, 1\rangle$, and let $\overrightarrow{c} = \langle 2, 1, 5\rangle$. It is known that $\overrightarrow{PQ} \times \overrightarrow{b} = \overrightarrow{0}$ and that $\overrightarrow{PQ} \cdot \overrightarrow{c} = 5$.
Find the point $P$.
I have no idea where to begin to solve this problem, but I do know the generic forms of the dot product and cross product, as so:
Generic equation for the cross product:
$$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$$
Generic equation for the dot product:
$$\overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$
So, using this knowledge we can derive an expression for $\overrightarrow{PQ}$, given the representation of point $P$ as $(p_1, p_2, p_3)$:
$$\overrightarrow{PQ} = \langle 1-p_1, 2-p_2, 3-p_3 \rangle$$
Then, in order to satisfy $\overrightarrow{PQ} \cdot \overrightarrow{c} = 5$, we can write the expression:
$$\langle 2, 1, 5 \rangle \cdot \langle 1-p_1, 2-p_2, 3-p_3 \rangle = (2-2p_1) + (2-p_2) + (15-5p_3) = 5$$
But I'm not sure where to go from here. I assume I could solve for a generic vector in terms of $P$ and combine this function with the one using the cross product, in a similar fashion.
How can I find the value of point $P$ by combining the dot product and cross product formulas?
Thank you so much for your help; it is super appreciated!
If we consider the dot product result (which you provided), and cross product equation (which returns the result $\overrightarrow{PQ}×\overrightarrow{b}=<-p_1,2p_2,-p_3>$), we get a system of equations:
$$ 2p_1 + 2p_2 + 5p_3 = 14 \\ -p_1+2p_2-p_3=0 $$
after much algebraic manipulation, we get: $p_1=\frac{14}{3}, p_2=\frac{14}{6}, p_3=0$