Question

the direction cosines of a vector A' are cos alpha = 4/5root 2, cos beeta= 1/root 2 and cos gama= 3/5root2 then the vector A is​

Answer 1

Answer:

The vector A is $\vec A=4\hat i+5\hat j+3\hat k$.

Explanation:

Given the directional cosines of a vector A.

The directional cosines or direction cosine of a vector, $\vec A=a\hat i+b\hat j+c\hat k$ is defined as the cosines of the angle subtended by a vector with the three coordinate axes.  

And the angles given by

$cos \alpha =\dfrac{a}{\sqrt{a^{2} +b^{2}+c^{2} }},~cos \beta =\dfrac{b}{\sqrt{a^{2} +b^{2}+c^{2} }}~\&~cos \gamma = \dfrac{c}{\sqrt{a^{2} +b^{2}+c^{2} }}$

Given

$cos \alpha = \dfrac{4}{5\sqrt{2}} ,~ cos \beta=\dfrac{1}{\sqrt{2}} ~\&~cos \gamma = \dfrac{3}{5\sqrt{2}}$

In the three angles given, cosα and cosγ has the denominator $5\sqrt{2}$

whereas the denominator for cosβ is $\sqrt{2}$.

To equate it to $5\sqrt{2}$, multiplying the numerator and denominator with 5,

$cos \beta=\dfrac{5}{5\sqrt{2}}$

Thus, we have,

$cos \alpha = \dfrac{4}{5\sqrt{2}} ,~ cos \beta=\dfrac{5}{5\sqrt{2}} ~\&~cos \gamma = \dfrac{3}{5\sqrt{2}}$

Now, comparing the given cosα, cosβ and cosγ with the formulas, we get

$a = 4, b =5~\&~c=3$

Therefore, the vector can be written as $\vec A=4\hat i+5\hat j+3\hat k$.

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