Question

How far is the star \( V \) from star \( A \)? 
 

Answer 1

Step 1: Compute the position vector of \( \overrightarrow{AV} \)
\[ \overrightarrow{AV} = {Position vector of } V - {Position vector of } A \] \[ \overrightarrow{AV} = (-3\hat{i} + 7\hat{j} + 11\hat{k}) - (7\hat{i} + 5\hat{j} + 8\hat{k}) = -10\hat{i} + 2\hat{j} + 3\hat{k}. \] 
Step 2: Compute the magnitude of \( \overrightarrow{AV} \)
\[ |\overrightarrow{AV}| = \sqrt{(-10)^2 + 2^2 + 3^2} = \sqrt{100 + 4 + 9} = \sqrt{113}. \] 
Step 3: Final result
The distance between star \( V \) and star \( A \) is \( \sqrt{113} \) units.

Answer 2

Step 1: Compute \( \overrightarrow{DA} \)
\[ \overrightarrow{DA} = {Position vector of } A - {Position vector of } D \] \[ \overrightarrow{DA} = (7\hat{i} + 5\hat{j} + 8\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k}) = 5\hat{i} + 2\hat{j} + 4\hat{k}. \] Step 2: Find the magnitude of \( \overrightarrow{DA} \)
\[ |\overrightarrow{DA}| = \sqrt{(5)^2 + (2)^2 + (4)^2} = \sqrt{25 + 4 + 16} = \sqrt{45} = 3\sqrt{5}. \] Step 3: Compute the unit vector
The unit vector is: \[ \hat{u} = \frac{\overrightarrow{DA}}{|\overrightarrow{DA}|} = \frac{5\hat{i} + 2\hat{j} + 4\hat{k}}{3\sqrt{5}} = \frac{5}{3\sqrt{5}}\hat{i} + \frac{2}{3\sqrt{5}}\hat{j} + \frac{4}{3\sqrt{5}}\hat{k}. \] Step 4: Final result
The unit vector in the direction of \( \overrightarrow{DA} \) is: \[ \frac{5}{3\sqrt{5}}\hat{i} + \frac{2}{3\sqrt{5}}\hat{j} + \frac{4}{3\sqrt{5}}\hat{k}. \]

Answer 3

Step 1: Recall the formula for the angle between vectors
The angle \( \theta \) between two vectors \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \) is given by: \[ \cos \theta = \frac{\overrightarrow{VD} \cdot \overrightarrow{DA}}{|\overrightarrow{VD}| \cdot |\overrightarrow{DA}|}. \] Step 2: Compute \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \)
From previous calculations: \[ \overrightarrow{VD} = \overrightarrow{V} - \overrightarrow{D} = (-3\hat{i} + 7\hat{j} + 11\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5\hat{i} + 4\hat{j} + 7\hat{k}. \] \[ \overrightarrow{DA} = 5\hat{i} + 2\hat{j} + 4\hat{k}. \] Step 3: Compute \( \overrightarrow{VD} \cdot \overrightarrow{DA} \)
\[ \overrightarrow{VD} \cdot \overrightarrow{DA} = (-5)(5) + (4)(2) + (7)(4) = -25 + 8 + 28 = 11. \] Step 4: Compute magnitudes of \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \)
\[ |\overrightarrow{VD}| = \sqrt{(-5)^2 + 4^2 + 7^2} = \sqrt{25 + 16 + 49} = \sqrt{90}. \] \[ |\overrightarrow{DA}| = \sqrt{(5)^2 + (2)^2 + (4)^2} = \sqrt{25 + 4 + 16} = \sqrt{45}. \] Step 5: Compute \( \cos \theta \)
\[ \cos \theta = \frac{11\sqrt{2}}{\sqrt{90} \cdot \sqrt{45}} = \frac{11\sqrt{2}}{\sqrt{4050}} = \frac{11\sqrt{2}}{90}. \] Step 6: Final result
The measure of \( \angle VDA \) is: \[ \theta = \cos^{-1} \left( \frac{11\sqrt{2}}{90} \right). \]

Answer 4

Step 1: Recall the formula for projection
The projection of \( \overrightarrow{DV} \) on \( \overrightarrow{DA} \) is given by: \[ {Projection} = \frac{\overrightarrow{DV} \cdot \overrightarrow{DA}}{|\overrightarrow{DA}|}. \] 
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
\[ {Projection} = \frac{\overrightarrow{DV} \cdot \overrightarrow{DA}}{|\overrightarrow{DA}|} = \frac{11}{3\sqrt{5}}. \] 
Step 6: Final result
The projection of \( \overrightarrow{DV} \) on \( \overrightarrow{DA} \) is: \[ \frac{11\sqrt{5}}{15}. \]