Question

What is the projection of vector \( \overrightarrow{DV} \) on vector \( \overrightarrow{DA} \)?

Hint:

The projection of one vector onto another gives the component of the first vector along the direction of the second.

Answer 1

Step 1: Recall the formula for projection
The projection of \( \overrightarrow{DV} \) on \( \overrightarrow{DA} \) is given by: \[ \text{Projection} = \frac{\overrightarrow{DV} \cdot \overrightarrow{DA}}{|\overrightarrow{DA}|}. \] Step 2: Compute \( \overrightarrow{DV} \)
\[ \overrightarrow{DV} = \overrightarrow{V} - \overrightarrow{D} = (-5\hat{i} + 4\hat{j} + 7\hat{k}). \] Step 3: Compute \( \overrightarrow{DV} \cdot \overrightarrow{DA} \)
From the previous calculations: \[ \overrightarrow{DV} \cdot \overrightarrow{DA} = (-5)(5) + (4)(2) + (7)(4) = -25 + 8 + 28 = 11. \] Step 4: Compute \( |\overrightarrow{DA}| \)
\[ |\overrightarrow{DA}| = \sqrt{(5)^2 + (2)^2 + (4)^2} = \sqrt{45} = 3\sqrt{5}. \] Step 5: Compute the projection
\[ \text{Projection} = \frac{\overrightarrow{DV} \cdot \overrightarrow{DA}}{|\overrightarrow{DA}|} = \frac{11}{3\sqrt{5}}. \] Step 6: Final result
The projection of \( \overrightarrow{DV} \) on \( \overrightarrow{DA} \) is: \[ \frac{11\sqrt{5}}{15}. \]