Write two different vectors having the same direction.
(N/A) Consider $\vec{p} = (\hat{i} + \hat{j} + \hat{k})$ and $\vec{q} = (2\hat{i} + 2\hat{j} + 2\hat{k})$.
The direction cosines of $\vec{p}$ are given by:
$l = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$,$m = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$,$n = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$.
The direction cosines of $\vec{q}$ are given by:
$l = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$m = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$n = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Since the direction cosines of $\vec{p}$ and $\vec{q}$ are the same,the two vectors have the same direction.
The direction cosines of $\vec{p}$ are given by:
$l = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$,$m = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$,$n = \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}$.
The direction cosines of $\vec{q}$ are given by:
$l = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$m = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$n = \frac{2}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Since the direction cosines of $\vec{p}$ and $\vec{q}$ are the same,the two vectors have the same direction.
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View SolutionLet $\vec{i}+\vec{j}+\vec{k}$, $a_1 \vec{i}+b_1 \vec{j}+c_1 \vec{k}$, $a_2 \vec{i}+b_2 \vec{j}+c_2 \vec{k}$, and $a_3 \vec{i}+b_3 \vec{j}+c_3 \vec{k}$ be the position vectors of the points $A, B, C, D$ respectively. The position vector of the centroid of the triangular face $BCD$ is $\frac{2}{3}(\vec{i}+\vec{j}+\vec{k})$. If $\alpha \vec{i}+\beta \vec{j}+\gamma \vec{k}$ is the position vector of the centroid of the tetrahedron $ABCD$, then find the value of $2 \alpha+\beta+\gamma$.
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