Question

Find the angle between the vectors \[ \vec{a}=\hat{i}+\hat{j}-\hat{k} \] and \[ \vec{b}=\hat{i}-\hat{j}+\hat{k}. \]

• A. \(0^\circ\)
• B. \(90^\circ\)
• C. \(\cos^{-1}\left(-\frac13\right)\)
• D. \(\cos^{-1}\left(\frac13\right)\)

Hint:

If the dot product is negative, the angle between the vectors is obtuse.

Answer 1

The Correct Option is C

Concept: The angle between two vectors can be found using the dot product formula: \[ \vec a \cdot \vec b = |\vec a|\,|\vec b|\cos\theta \] Therefore, \[ \cos\theta = \frac{\vec a\cdot\vec b} {|\vec a||\vec b|} \] where \[ \theta = \text{angle between the vectors}. \]

Step 1: Write the given vectors \[ \vec a=(1,1,-1) \] \[ \vec b=(1,-1,1) \]

Step 2: Find the dot product Using \[ \vec a\cdot\vec b = a_1b_1+a_2b_2+a_3b_3 \] we get \[ =(1)(1)+(1)(-1)+(-1)(1) \] \[ =1-1-1 \] \[ =-1 \] Hence, \[ \vec a\cdot\vec b=-1 \]

Step 3: Find magnitude of each vector For vector \(\vec a\), \[ |\vec a| = \sqrt{1^2+1^2+(-1)^2} \] \[ = \sqrt{1+1+1} \] \[ = \sqrt3 \] Similarly, \[ |\vec b| = \sqrt{1^2+(-1)^2+1^2} \] \[ = \sqrt3 \]

Step 4: Apply the angle formula \[ \cos\theta = \frac{-1} {\sqrt3\times\sqrt3} \] \[ = -\frac13 \] Therefore, \[ \theta = \cos^{-1} \left(-\frac13\right) \] Final Answer: \[ \boxed{\cos^{-1}\left(-\frac13\right)} \]