Question

If vector a = 2i+3j+6k and b=3i+4j then find projection of a on b also find projection of b on a​

Answer 1

Answer:

Answer:

The projection of \vec{a}

a

on \vec{b}

b

is 3.

The projection of \vec{b}

b

on \vec{a}

a

is \frac{15}{7}

7

15

.

Step-by-step explanation:

Given : If vector a = 2i+3j+6k and b=3i+4j.

To find : Projection of a on b also projection of b on a?

Solution :

The formula of projection of \vec{a}

a

on \vec{b}

b

is =\frac{1}{|\vec{b}|}(\vec{a}\cdot\vec{b})=

∣

b

∣

1

(

a

⋅

b

)

a = 2i+3j+6k and b=3i+4j

\vec{a}\cdot\vec{b}=2(3)+3(3)+6(0)

a

⋅

b

=2(3)+3(3)+6(0)

\vec{a}\cdot\vec{b}=6+9+0

a

⋅

b

=6+9+0

\vec{a}\cdot\vec{b}=15

a

⋅

b

=15

Magnitude of b,

|\vec{b}|=\sqrt{3^2+4^2}∣

b

∣=

3

2

+4

2

|\vec{b}|=\sqrt{9+16}∣

b

∣=

9+16

|\vec{b}|=\sqrt{25}∣

b

∣=

25

|\vec{b}|=5∣

b

∣=5

Substitute the value in the formula,

Projection of \vec{a}

a

on \vec{b}

b

is =\frac{1}{|\vec{b}|}(\vec{a}\cdot\vec{b})=

∣

b

∣

1

(

a

⋅

b

)

=\frac{1}{5}(15)=

5

1

(15)

=3=3

The projection of \vec{a}

a

on \vec{b}

b

is 3.

The formula of projection of \vec{b}

b

on \vec{a}

a

is =\frac{1}{|\vec{a}|}(\vec{a}\cdot\vec{b})=

∣

a

∣

1

(

a

⋅

b

)

a = 2i+3j+6k and b=3i+4j

\vec{a}\cdot\vec{b}=15

a

⋅

b

=15

Magnitude of a,

|\vec{a}|=\sqrt{2^2+3^2+6^2}∣

a

∣=

2

2

+3

2

+6

2

|\vec{a}|=\sqrt{4+9+36}∣

a

∣=

4+9+36

|\vec{a}|=\sqrt{49}∣

a

∣=

49

|\vec{a}|=7∣

a

∣=7

Substitute the value in the formula,

Projection of \vec{b}

b

on \vec{a}

a

is =\frac{1}{|\vec{b}|}(\vec{a}\cdot\vec{b})=

∣

b

∣

1

(

a

⋅

b

)

=\frac{1}{7}(15)=

7

1

(15)

=\frac{15}{7}=

7

15

The projection of \vec{b}

b

on \vec{a}

a

is \frac{15}{7}

7

15

.

Explanation:

projection of a on b/projection of b on a = 3/15/7

-»= 5/7 is the answer....

hope that helps..

Answer 2

Answer:

The projections of vector a on b and vice versa are 3.6 and 2.57 respectively.

Explanation:

The given vectors are $a=2i+3j+6k\\b=3i+4j$

We are required to find the projection of the vector a on b and also the projection of the vector b on a.

For this we shall use an formula from vector algebra which is written as shown.

The projection of vector a on b is $\frac{a\vec\overrightarro.b\vec\overrightarro}{|b\vec\overrightarro|}$

Let us find the magnitude of both the vectors as follows-

$|a\vec\overrightarro|=\sqrt{2^{2}+ 3^{2}+ 6^{2} } =\sqrt{49} =7\\|b\vec\overrightarro|=\sqrt{ 3^{2}+ 4^{2} } =\sqrt{25} =5$

We shall also find the dot product of both the vectors as follows-

$a\vec\overrightarro.b\vec\overrightarro=(2i+3j+6k).(3i+4j)=2\times3+3\times4=18$

Hence the projection of vector a on b is $\frac{18}{5} =3.6$

Similarly,The projection of vector b on a is $\frac{a\vec\overrightarro.b\vec\overrightarro}{|a\vec\overrightarro|}$an it is equal to $\frac{18}{7} =2.57$

Therefore,The projections of vector a on b and vice versa are 3.6 and 2.57 respectively.

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