Source: Extra Practice

If a unit vector is represented by 0.5i^0.8j^+ck^0.5\hat{i} - 0.8\hat{j} + c\hat{k}, then the value of cc is:

Options

Option A is correct

0.11\sqrt{0.11}

Option B

0.39\sqrt{0.39}

Option C

0.010.01

Option D

0.01\sqrt{0.01}

Explanation

A unit vector has a magnitude of exactly 11. Therefore, the magnitude of the given vector is (0.5)2+(0.8)2+c2=1\sqrt{(0.5)^2 + (-0.8)^2 + c^2} = 1. Squaring both sides gives 0.25+0.64+c2=1    0.89+c2=1    c2=10.89=0.11    c=0.110.25 + 0.64 + c^2 = 1 \implies 0.89 + c^2 = 1 \implies c^2 = 1 - 0.89 = 0.11 \implies c = \sqrt{0.11}. The other options incorrect because they do not satisfy the unit magnitude condition: 0.25+0.64+0.3910.25 + 0.64 + 0.39 \neq 1, 0.25+0.64+0.01210.25 + 0.64 + 0.01^2 \neq 1, and 0.25+0.64+0.0110.25 + 0.64 + 0.01 \neq 1.