Question

If A =2i +4j-5k the direction of cosines of the vector A

Answer 1

Answer:

Explanation:

Direction cosines of a vector A=cos alpha, cos beta, cos gama where

Cos alpha= x/|A|= 2/45^1/2

Cos beta =y/|A|=4/45^1/2

Cos gama=z/|A|=-5/45^1/2

Answer 2

Answer:

The directions of the cosines of the vector is $\frac{2}{\sqrt{45}}$,$\frac{4}{\sqrt{45}}$ and $\frac{-5}{\sqrt{45}}$.

Explanation:

The direction of the cosine of a vector is determined by dividing the corresponding coordinate of a vector by the vector length. The unit vector coordinate is equal to the direction of the cosine. Suppose a vector is given as

$\vec A=a\hat{i}+b\hat{j}+c\hat{k}$        (1)

The directions of cosines of the vector are given as;

$cos\alpha=\frac{a}{\sqrt{a^2+b^2+c^2}}$     (2)

$cos\beta=\frac{b}{\sqrt{a^2+b^2+c^2}}$     (3)

$cos\gamma=\frac{a}{\sqrt{a^2+b^2+c^2}}$      (4)

The vector given in the question is,

$\vec A=2\hat{i}+4\hat{j}-5\hat{k}$       (5)

The direction of cosines are,

$cos\alpha=\frac{2}{\sqrt{2^2+4^2+(-5)^2}}=\frac{2}{\sqrt{45}}$      (6)

$cos\beta=\frac{4}{\sqrt{2^2+4^2+(-5)^2}}=\frac{4}{\sqrt{45}}$     (7)

$cos\gamma=\frac{-5}{\sqrt{2^2+4^2+(-5)^2}}=\frac{-5}{\sqrt{45}}$     (8)

Hence, the directions of the cosines of the vector is $\frac{2}{\sqrt{45}}$,$\frac{4}{\sqrt{45}}$ and $\frac{-5}{\sqrt{45}}$.