What is the sum of the squares of direction cosines of the line joining the points `(1,2,-3) and (-2,3,1)`?

Options

A. 0
B. 1
C. 3
D. `2/sqrt(26)`

Answer

To find the sum of the squares of the direction cosines of the line joining the points \( (1, 2, -3) \) and \( (-2, 3, 1) \), we can follow these steps:

### Step 1: Find the direction vector of the line
The direction vector \( \vec{d} \) of the line joining the two points \( A(1, 2, -3) \) and \( B(-2, 3, 1) \) can be calculated as:
\[ \vec{d} = B - A = (-2 - 1, 3 - 2, 1 - (-3)) = (-3, 1, 4) \]

### Step 2: Calculate the magnitude of the direction vector
The magnitude \( |\vec{d}| \) of the direction vector \( \vec{d} = (-3, 1, 4) \) is given by:
\[ |\vec{d}| = \sqrt{(-3)^2 + 1^2 + 4^2} = \sqrt{9 + 1 + 16} = \sqrt{26} \]

### Step 3: Find the direction cosines
The direction cosines \( \cos \alpha, \cos \beta, \cos \gamma \) can be found using the components of the direction vector divided by its magnitude:
\[ \cos \alpha = \frac{-3}{\sqrt{26}}, \quad \cos \beta = \frac{1}{\sqrt{26}}, \quad \cos \gamma = \frac{4}{\sqrt{26}} \]

### Step 4: Calculate the sum of the squares of the direction cosines
Now, we calculate the sum of the squares of the direction cosines:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \left(\frac{-3}{\sqrt{26}}\right)^2 + \left(\frac{1}{\sqrt{26}}\right)^2 + \left(\frac{4}{\sqrt{26}}\right)^2 \]
\[ = \frac{9}{26} + \frac{1}{26} + \frac{16}{26} = \frac{9 + 1 + 16}{26} = \frac{26}{26} = 1 \]

### Conclusion
The sum of the squares of the direction cosines of the line joining the points \( (1, 2, -3) \) and \( (-2, 3, 1) \) is \( 1 \).

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