1. Prove that OP = (5/13, 12/13) is a unit vector. 2. If P(2, 3) and Q(2, 3) are two points, prove that PQ is a null vector.
Understand the Problem
The image contains two math proof questions related to vectors: The first question asks to prove that vector OP is a unit vector. The second asks to prove that vector PQ is a null vector given points P and Q.
Answer
$|\vec{OP}| = 1$, therefore $\vec{OP}$ is a unit vector $\vec{PQ} = \binom{0}{0}$, therefore $\vec{PQ}$ is a null vector
Answer for screen readers
$\vec{OP}$ is a unit vector because $|\vec{OP}| = 1$.
$\vec{PQ}$ is a null vector because $\vec{PQ} = \binom{0}{0}$.
Steps to Solve
- Prove $\vec{OP}$ is a unit vector
A unit vector has a magnitude (or length) of 1. To prove that $\vec{OP}$ is a unit vector, we must show that its magnitude is equal to 1. We are given $\vec{OP} = \binom{5/13}{12/13}$. The magnitude of a vector $\binom{a}{b}$ is $\sqrt{a^2 + b^2}$.
Compute the magnitude of $\vec{OP}$:
$$|\vec{OP}| = \sqrt{\left(\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2}$$
- Simplify the magnitude
$$|\vec{OP}| = \sqrt{\frac{25}{169} + \frac{144}{169}} = \sqrt{\frac{169}{169}} = \sqrt{1} = 1$$
Since the magnitude of $\vec{OP}$ is 1, $\vec{OP}$ is a unit vector.
- Prove $\vec{PQ}$ is a null vector
A null vector (or zero vector) has all components equal to zero. We are given points $P(2, 3)$ and $Q(2, 3)$. The vector $\vec{PQ}$ can be found by subtracting the coordinates of point $P$ from the coordinates of point $Q$. That is: $\vec{PQ} = Q - P$.
- Calculate $\vec{PQ}$
$$\vec{PQ} = \binom{2}{3} - \binom{2}{3} = \binom{2-2}{3-3} = \binom{0}{0}$$
Since $\vec{PQ} = \binom{0}{0}$, $\vec{PQ}$ is a null vector
$\vec{OP}$ is a unit vector because $|\vec{OP}| = 1$.
$\vec{PQ}$ is a null vector because $\vec{PQ} = \binom{0}{0}$.
More Information
A unit vector has a magnitude of 1, while a null vector has all components equal to zero.
Tips
Null.
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