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If a\vec{a} and b\vec{b} are two non-zero orthogonal vectors, then a+b|\vec{a} + \vec{b}| is equal to

Solution

Correct Option: 4

Two vectors a\vec{a} and b\vec{b} are non-zero and orthogonal. When two vectors are orthogonal, their dot product is zero: ab=0\vec{a} \cdot \vec{b} = 0.


To find the magnitude of the vector sum:

a+b2=(a+b)(a+b)|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})

=aa+ab+ba+bb= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}

=a2+2ab+b2= |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2

Since the vectors are orthogonal, ab=0\vec{a} \cdot \vec{b} = 0:

a+b2=a2+b2|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2

a+b=a2+b2|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2}


Similarly, for ab|\vec{a} - \vec{b}|:

ab2=(ab)(ab)|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})

=a22ab+b2= |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{b}|^2

Since ab=0\vec{a} \cdot \vec{b} = 0:

ab2=a2+b2|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2

ab=a2+b2|\vec{a} - \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2}


Comparing the results:

a+b=a2+b2=ab|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = |\vec{a} - \vec{b}|

Therefore, a+b=ab|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|

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