Question

Find p such that the vectors w=pi+3j and v=2i + qj are parallel to u=5i + 6j.

2.7
2.5
3.5
4.1

Answer 1

Answer on Question #46935 – Math – Vector Calculus

Question.

Find $p$ such that the vectors $w = pi + 3j$ and $v = 2i + qj$ are parallel to $u = 5i + 6j$.

2.7

2.5

3.5

4.1

Solution.

By definition, the magnitude of the cross product is calculated by the following formula:

$$\left| \vec {a} \times \vec {b} \right| = | \vec {a} | | \vec {b} | \sin \alpha$$

So, if vectors $\vec{a}$ and $\vec{b}$ are parallel $\rightarrow \alpha = \pi n \rightarrow \sin \alpha = 0 \rightarrow \vec{a} \times \vec{b} = \vec{0}$.

Therefore, we must use the condition $\vec{w} \times \vec{u} = \vec{0}$.

In our case, the coordinates of vectors are the following:

$$\vec {w} = (p; 3; 0)$$ $$\vec {u} = (5; 6; 0)$$

Let find the value of $p$ using the condition $\vec{w} \times \vec{u} = \vec{0}$:

$$\vec {w} \times \vec {u} = \left| \begin{array}{c c c} \vec {i} & \vec {j} & \vec {k} \\ p & 3 & 0 \\ 5 & 6 & 0 \end{array} \right| = \vec {i} \cdot 0 + \vec {j} \cdot 0 + \vec {k} \cdot (6 p - 1 5) = \vec {k} \cdot (6 p - 1 5) = 0 \rightarrow p = 2. 5$$

Answer.

2.5

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