What is the projection of ## onto ##?

Question

What is the projection of #<6,5,3 ># onto #<2,-1,8 >#?

Charlotte Aaron · Accepted Answer

The vector projection is #< 62/69,-31/69,248/69 >#, the scalar projection is #(31sqrt(69))/69#.

Given #veca= < 6,5,3 ># and #vecb= < 2,-1,8 >#, we can find #proj_(vecb)veca#, the vector projection of #veca# onto #vecb# using the following formula: #proj_(vecb)veca=((veca*vecb)/(|vecb|))vecb/|vecb|# That is, the dot product of the two vectors divided by the magnitude of #vecb#, multiplied by #vecb# divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide #vecb# by its magnitude in order to obtain a unit vector (vector with magnitude of #1#). You might notice that the first quantity is scalar, as we know that when we take the dot product of two vectors, the resultant is a scalar. Therefore, the scalar projection of #a# onto #b# is #comp_(vecb)veca=(a*b)/(|b|)#, also written #|proj_(vecb)veca|#. We can start by taking the dot product of the two vectors. #veca*vecb=< 6,5,3 > * < 2,-1,8 ># #=> (6*2)+(5*-1)+(3*8)# #=>12-5+24=31# Then we can find the magnitude of #vecb# by taking the square root of the sum of the squares of each of the components. #|vecb|=sqrt((b_x)^2+(b_y)^2+(b_z)^2)# #|vecb|=sqrt((2)^2+(-1)^2+(8)^2)# #=>sqrt(4+1+64)=sqrt(69)# And now we have everything we need to find the vector projection of #veca# onto #vecb#. #proj_(vecb)veca=(31)/sqrt(69)*(< 2,-1,8 >)/sqrt(69)# #=>(31 < 2,-1,8 >)/69# #=31/69< 2,-1,8 ># You can distribute the coefficient to each component of the vector and write as: #=>< 62/69,-31/69,248/69 ># The scalar projection of #veca# onto #vecb# is just the first half of the formula, where #comp_(vecb)veca=(a*b)/(|b|)#. Therefore, the scalar projection is #31/sqrt(69)#, which does not simplify any further, besides to rationalize the denominator if desired, giving #(31sqrt(69))/69#. Hope that helps!

Adam Arnold · Answer

undefined To find the projection of vector $ \mathbf{v} = \langle 6, 5, 3 \rangle $ onto vector $ \mathbf{u} = \langle 2, -1, 8 \rangle $, we use the formula:

$$
\text{proj}_{\mathbf{u}}(\mathbf{v}) = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{\| \mathbf{u} \| ^2} \right) \mathbf{u}
$$

where $ \cdot $ represents the dot product and $ \| \mathbf{u} \| $ represents the magnitude of vector $ \mathbf{u} $.

First, calculate the dot product:

$$
\mathbf{v} \cdot \mathbf{u} = (6 \cdot 2) + (5 \cdot -1) + (3 \cdot 8) = 12 - 5 + 24 = 31
$$

Then, find the magnitude of $ \mathbf{u} $:

$$
\| \mathbf{u} \| = \sqrt{2^2 + (-1)^2 + 8^2} = \sqrt{4 + 1 + 64} = \sqrt{69}
$$

Now, substitute these values into the formula:

$$
\text{proj}_{\mathbf{u}}(\mathbf{v}) = \left( \frac{31}{69} \right) \langle 2, -1, 8 \rangle = \langle \frac{62}{69}, -\frac{31}{69}, \frac{248}{69} \rangle
$$

So, the projection of $ \langle 6, 5, 3 \rangle $ onto $ \langle 2, -1, 8 \rangle $ is approximately $ \langle \frac{62}{69}, -\frac{31}{69}, \frac{248}{69} \rangle $.

What is the projection of #&lt;6,5,3 &gt;# onto #&lt;2,-1,8 &gt;#?

What is the projection of #<6,5,3 ># onto #<2,-1,8 >#?