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If a\vec{a} and b\vec{b} are two unit vectors and a+b\vec{a} + \vec{b} is also unit vector, the magnitude of ab\vec{a} - \vec{b} is

Solution

Correct Option: 4

Given:

  • a=1|\vec{a}| = 1 (unit vector)
  • b=1|\vec{b}| = 1 (unit vector)
  • a+b=1|\vec{a} + \vec{b}| = 1 (unit vector)

For the magnitude of a vector sum:

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

Since a+b=1|\vec{a} + \vec{b}| = 1, a=1|\vec{a}| = 1, and b=1|\vec{b}| = 1:

1=1+1+2ab1 = 1 + 1 + 2\vec{a} \cdot \vec{b}

1=2+2ab1 = 2 + 2\vec{a} \cdot \vec{b}

2ab=12\vec{a} \cdot \vec{b} = -1

ab=12\vec{a} \cdot \vec{b} = -\frac{1}{2}


For the magnitude of a vector difference:

ab2=a2+b22ab|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}

ab2=1+12(12)|\vec{a} - \vec{b}|^2 = 1 + 1 - 2\left(-\frac{1}{2}\right)

ab2=2+1|\vec{a} - \vec{b}|^2 = 2 + 1

ab2=3|\vec{a} - \vec{b}|^2 = 3

ab=3|\vec{a} - \vec{b}| = \sqrt{3}

Therefore, the magnitude of ab\vec{a} - \vec{b} is 3\sqrt{3}.

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