Question:
If $\alpha = (-1,0,1), \beta = (2,0,-2) \in V_{3}(R)$ an inner product space, then $||\alpha + \beta|| = $
If $\alpha = (-1,0,1), \beta = (2,0,-2) \in V_{3}(R)$ an inner product space, then $||\alpha + \beta|| = $
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Norm is just the 3D version of the Pythagorean theorem. Always sum the components before taking the squares.
Updated On: Jul 3, 2026
- 2
- $\sqrt{2}$
- $-\sqrt{2}$
- $\sqrt{2}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The norm (length) of a vector $v = (x, y, z)$ is $||v|| = \sqrt{x^2 + y^2 + z^2}$.
Step 2: Meaning
First, find the sum vector $\alpha + \beta = (-1+2, 0+0, 1-2) = (1, 0, -1)$.
Step 3: Analysis
Now calculate the norm of $(1, 0, -1)$: $\sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1}$.
Step 4: Conclusion
This equals $\sqrt{2}$. Final Answer: (D)
The norm (length) of a vector $v = (x, y, z)$ is $||v|| = \sqrt{x^2 + y^2 + z^2}$.
Step 2: Meaning
First, find the sum vector $\alpha + \beta = (-1+2, 0+0, 1-2) = (1, 0, -1)$.
Step 3: Analysis
Now calculate the norm of $(1, 0, -1)$: $\sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1}$.
Step 4: Conclusion
This equals $\sqrt{2}$. Final Answer: (D)
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