Question

If $A = (-2, 2, 3)$, $B = (3, 2, 2)$, $C = (4, -3, 5)$, and $D = (7, -5, -1)$, then the projection of $\vec{AB}$ on $\vec{CD}$ is

• A. $4$
• B. $3$
• C. $\frac{12}{7}$
• D. None of these

Hint:

Always double check which vector is the base of the projection. The magnitude of whichever vector follows the word "on" must always be placed alone in the denominator!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given four coordinates in 3D space: $A$, $B$, $C$, and $D$. We need to construct vectors $\vec{AB}$ and $\vec{CD}$ and compute the scalar projection of vector $\vec{AB}$ onto vector $\vec{CD}$.

Step 2: Key Formula or Approach:
The scalar projection of a vector $\vec{u}$ onto another vector $\vec{v}$ is given by the formula: $$\text{Projection of } \vec{u} \text{ on } \vec{v} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$$ Here, $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{CD}$.

Step 3: Detailed Explanation:
First, let's find the components of vectors $\vec{AB}$ and $\vec{CD}$: $$\vec{AB} = (3 - (-2))\hat{i} + (2 - 2)\hat{j} + (2 - 3)\hat{k} = 5\hat{i} + 0\hat{j} - \hat{k}$$ $$\vec{CD} = (7 - 4)\hat{i} + (-5 - (-3))\hat{j} + (-1 - 5)\hat{k} = 3\hat{i} - 2\hat{j} - 6\hat{k}$$ Now compute the dot product $\vec{AB} \cdot \vec{CD}$: $$\vec{AB} \cdot \vec{CD} = (5)(3) + (0)(-2) + (-1)(-6) = 15 + 0 + 6 = 21$$ Next, find the magnitude of the target base vector $\vec{CD}$: $$|\vec{CD}| = \sqrt{3^2 + (-2)^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$ Substitute the dot product and magnitude into our projection formula: $$\text{Projection} = \frac{21}{7} = 3$$ This matches option (B).

Step 4: Final Answer:
The scalar projection of $\vec{AB}$ on $\vec{CD}$ is $3$, which corresponds to option (B).