What is the projection of $< 0, 1, 3 >$ $<0, 1, 3>$ onto $< 0, 4, 4 >$ $<0, 4, 4>$ ?

Question

What is the projection of $< 0, 1, 3 >$ $<0, 1, 3>$ onto $< 0, 4, 4 >$ $<0, 4, 4>$ ?

1 Answer

Impact of this question

1797 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-24T17:38:06

The vector projection is $< 0, 2, 2 >$ $< 0,2,2 >$ , the scalar projection is $2 \sqrt{2}$ $2sqrt2$ . See below.

Explanation:

Given $\to a = < 0, 1, 3 >$ $veca=< 0,1,3 >$ and $\to b = < 0, 4, 4 >$ $vecb= < 0,4,4 >$ , we can find $p r o j_{\to b} \to a$ $proj_(vecb)veca$ , the vector projection of $\to a$ $veca$ onto $\to b$ $vecb$ using the following formula:

$p r o j_{\to b} \to a = ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\to a \cdot \to b}{∣ ∣ ∣ \to b ∣ ∣ ∣} ⎞ ⎟ ⎟ ⎟ ⎠ \frac{\to b}{∣ ∣ ∣ \to b ∣ ∣ ∣}$ $proj_(vecb)veca=((veca*vecb)/(|vecb|))vecb/|vecb|$

That is, the dot product of the two vectors divided by the magnitude of $\to b$ $vecb$ , multiplied by $\to b$ $vecb$ divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide $\to b$ $vecb$ by its magnitude in order to obtain a unit vector (vector with magnitude of $1$ $1$ ). 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 $a$ $a$ onto $b$ $b$ is $c o m p_{\to b} \to a = \frac{a \cdot b}{| b |}$ $comp_(vecb)veca=(a*b)/(|b|)$ , also written $∣ ∣ p r o j_{\to b} \to a ∣ ∣$ $|proj_(vecb)veca|$ .

We can start by taking the dot product of the two vectors:

$\to a \cdot \to b = < 0, 1, 3 > \cdot < 0, 4, 4 >$ $veca*vecb=< 0,1,3 >*< 0,4,4 >$

$\Rightarrow (0 \cdot 0) + (4 \cdot 1) + (4 \cdot 3)$ $=> (0*0)+(4*1)+(4*3)$

$\Rightarrow 0 + 4 + 12 = 16$ $=>0+4+12=16$

Then we can find the magnitude of $\to b$ $vecb$ by taking the square root of the sum of the squares of each of the components.

$∣ ∣ ∣ \to b ∣ ∣ ∣ = \sqrt{{(b_{x})}_{x}^{2} + {(b_{y})}_{y}^{2} + {(b_{z})}_{z}^{2}}$ $|vecb|=sqrt((b_x)^2+(b_y)^2+(b_z)^2)$

$∣ ∣ ∣ \to b ∣ ∣ ∣ = \sqrt{{(0)}^{2} + {(4)}^{2} + {(4)}^{2}}$ $|vecb|=sqrt((0)^2+(4)^2+(4)^2)$

$\Rightarrow \sqrt{0 + 16 + 16} = \sqrt{32}$ $=>sqrt(0+16+16)=sqrt(32)$

And now we have everything we need to find the vector projection of $\to a$ $veca$ onto $\to b$ $vecb$ .

$p r o j_{\to b} \to a = \frac{16}{\sqrt{32}} \cdot \frac{< 0, 4, 4 >}{\sqrt{32}}$ $proj_(vecb)veca=(16)/sqrt(32)*(< 0,4,4 >)/sqrt(32)$

$\Rightarrow \frac{16 < 0, 4, 4 >}{32}$ $=>(16 < 0,4,4 >)/32$

$\Rightarrow \frac{< 0, 4, 4 >}{2}$ $=>(< 0,4,4 >)/2$

$\Rightarrow < 0, 2, 2 >$ $=>< 0,2,2 >$

The scalar projection of $\to a$ $veca$ onto $\to b$ $vecb$ is just the first half of the formula, where $c o m p_{\to b} \to a = \frac{a \cdot b}{| b |}$ $comp_(vecb)veca=(a*b)/(|b|)$ . Therefore, the scalar projection is $\frac{16}{\sqrt{32}}$ $16/sqrt(32)$ , which further simplifies to $2 \sqrt{2}$ $2sqrt2$ . I've shown the simplification below.

$\frac{16}{\sqrt{32}}$ $16/sqrt(32)$

$\Rightarrow \frac{16}{\sqrt{16 \cdot 2}}$ $=>16/sqrt(16*2)$

$\Rightarrow \frac{16}{4 \cdot \sqrt{2}}$ $=>16/(4*sqrt2)$

$\Rightarrow \frac{4}{\sqrt{2}}$ $=>4/sqrt2$

$\Rightarrow \frac{4 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$ $=>(4*sqrt2)/(sqrt2*sqrt2)$

$\Rightarrow \frac{4 \sqrt{2}}{2}$ $=>(4sqrt2)/2$

$\Rightarrow 2 \sqrt{2}$ $=>2sqrt2$

Hope that helps!

Science

Math

Humanities

... and beyond

What is the projection of $< 0, 1, 3 >$ $<0, 1, 3>$ onto $< 0, 4, 4 >$ $<0, 4, 4>$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the projection of <0,1,3><0, 1, 3> onto <0,4,4><0, 4, 4>?

1 Answer

Explanation:

Related questions

Impact of this question

What is the projection of $< 0, 1, 3 >$ $<0, 1, 3>$ onto $< 0, 4, 4 >$ $<0, 4, 4>$ ?