If 0.5 hat{i} + 0.8 hat{j} + c hat{k} is a un | vedclass

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Question

(D) vector $\vec{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$ is a unit vector if its magnitude is $1$,i.e.,$|\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2}

Vedclass · Accepted Answer

$\sqrt{0.11}$

Vedclass · Answer

(D) vector $\vec{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$ is a unit vector if its magnitude is $1$,i.e.,$|\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = 1$.Given the vector is $0.5 \hat{i} + 0.8 \hat{j} + c \hat{k}$.Therefore,$\sqrt{(0.5)^2 + (0.8)^2 + c^2} = 1$.Squaring both sides,we get $(0.5)^2 + (0.8)^2 + c^2 = 1^2$.$0.25 + 0.64 + c^2 = 1$.$0.89 + c^2 = 1$.$c^2 = 1 - 0.89 = 0.11$.Thus,$c = \sqrt{0.11}$.

If $0.5 \hat{i} + 0.8 \hat{j} + c \hat{k}$ is a unit vector,then $c$ is

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