Question

obtain the direction cosines of vector a-b if a=2i+3j+k b=2i+2j+3k. ................don't spam please ​

Answer 1

Answer:

a - b = j - 2k

so, |a-b| = √(1+4) =√5

direction cosines ,

l = 0/√5=0

m= 1/√5

n = -2/√5

hence the direction cosines are 0 , 1/√5 , (-2)/√5.

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Answer 2

As per the data given in the above question.

We have to find the direction cosine of A-B

Given,

$A=2 \hat i+3 \hat j+\hat k$

$B=2 \hat i+2 \hat j+3 \hat k$

Explanation:

$Let \: \hat c \: be \: \hat A-\hat B$

$\hat c \: = \hat A-\hat B$

$\hat c= (2 \hat i+3 \hat j+\hat k)-(2 \hat i+2 \hat j+3 \hat k)$

$\hat c= (2 \hat i+3 \hat j+\hat k-2 \hat i - 2 \hat j - 3 \hat k)$

$\hat c= 0 \hat i + \hat j - 2\hat k$

Direction ratios are

$a=0 ,b= 1 \: and \: c= -2$

Magnitude ,

$|\hat c | = \sqrt{ {0}^{2} + {1}^{2} + { - 2}^{2} }$

$|\hat c | = \sqrt{0 + 1 + 4}$

$|\hat c | = \sqrt{5}$

Direction cosines are ,

$\frac{a}{|\hat c | } \frac{b}{|\hat c | } \: \frac{c}{|\hat c | }$

so ,

$\frac{a}{ \sqrt{5} } \frac{b}{ \sqrt{5} } \frac{c}{ \sqrt{5} }$

Hence,

$Direction \: cosines \: are \: \frac{a}{ \sqrt{5} } , \frac{b}{ \sqrt{5} } , \frac{c}{ \sqrt{5} } </p><p>$

Project code #SPJ2