If P→=i^+2j^+6k^ , its direction cosines are

P¯=i+2j+6k P^=P→|P| |P¯|=12+22+62=1+4+36=41 P^=1i+2j+6k41=1i41+2j41+6k41 D.C=141,241,641 Also Check: physics formulas

If P→=i^+2j^+6k^ , its direction cosines are

P¯=i+2j+6k P^=P→|P| |P¯|=12+22+62=1+4+36=41 P^=1i+2j+6k41=1i41+2j41+6k41 D.C=141,241,641 Also Check: physics formulas

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If $\vec{P} = \hat{i} + 2 \hat{j} + 6 \hat{k}$ , its direction cosines are

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\frac{1}{41}, \frac{2}{41} and \frac{6}{41}

\frac{1}{\sqrt{41}}, \frac{2}{\sqrt{41}} and \frac{6}{\sqrt{41}}

\frac{3}{\sqrt{41}}, \frac{8}{\sqrt{41}}, \frac{7}{\sqrt{41}}

1, 2 and 6

answer is B.

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Detailed Solution

\bar{P} = i + 2 j + 6 k

$\hat{P} = \frac{\vec{P}}{| P |}$

$| \bar{P} | = \sqrt{1^{2} + 2^{2} + 6^{2}} = \sqrt{1 + 4 + 36} = \sqrt{41}$

$\hat{P} = \frac{1 i + 2 j + 6 k}{\sqrt{41}} = \frac{1 i}{\sqrt{41}} + \frac{2 j}{\sqrt{41}} + \frac{6 k}{\sqrt{41}}$

$D .C = \frac{1}{\sqrt{41}}, \frac{2}{\sqrt{41}}, \frac{6}{\sqrt{41}}$

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If P→=i^+2j^+6k^ , its direction cosines are

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If $\vec{P} = \hat{i} + 2 \hat{j} + 6 \hat{k}$ , its direction cosines are