What is the projection of $< 7, 1, 1 >$ $<7,1,1 >$ onto $< - 1, 6, - 8 >$ $<-1,6,-8 >$ ?

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What is the projection of $< 7, 1, 1 >$ $<7,1,1 >$ onto $< - 1, 6, - 8 >$ $<-1,6,-8 >$ ?

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Answer 1 · 2017-01-03T01:29:21

The vector projection is $< \frac{9}{101}, - \frac{54}{101}, \frac{72}{101} >$ $< 9/101,-54/101,72/101 >$ , the scalar projection is $\frac{- 9 \sqrt{101}}{101}$ $(-9sqrt(101))/101$ .

Explanation:

Given $\to a = < 7, 1, 1 >$ $veca= < 7,1,1 >$ and $\to b = < - 1, 6, - 8 >$ $vecb= < -1,6,-8 >$ , 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.

$\to a \cdot \to b = < 7, 1, 1 > \cdot < - 1, 6, - 8 >$ $veca*vecb=< 7,1,1 > * < -1,6,-8 >$

$\Rightarrow (7 \cdot - 1) + (1 \cdot 6) + (1 \cdot - 8)$ $=> (7*-1)+(1*6)+(1*-8)$

$\Rightarrow - 7 + 6 - 8 = - 9$ $=>-7+6-8=-9$

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.

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

$∣ ∣ ∣ \to b ∣ ∣ ∣ = \sqrt{{(- 1)}^{2} + {(6)}^{2} + {(- 8)}^{2}}$ $|vecb|=sqrt((-1)^2+(6)^2+(-8)^2)$

$\Rightarrow \sqrt{1 + 36 + 64} = \sqrt{101}$ $=>sqrt(1+36+64)=sqrt(101)$

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

$p r o j_{\to b} \to a = \frac{- 9}{\sqrt{101}} \cdot \frac{< - 1, - 6, - 8 >}{\sqrt{101}}$ $proj_(vecb)veca=(-9)/sqrt(101)*(< -1,-6,-8 >)/sqrt(101)$

$\Rightarrow \frac{- 9 < - 1, 6, - 8 >}{101}$ $=>(-9 < -1,6,-8 >)/101$

$= - \frac{9}{101} < - 1, 6, - 8 >$ $=-9/101< -1,6,-8 >$

You can distribute the coefficient to each component of the vector and write as:

$\Rightarrow < \frac{9}{101}, - \frac{54}{101}, \frac{72}{101} >$ $=>< 9/101,-54/101,72/101 >$

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 $- \frac{9}{\sqrt{101}}$ $-9/sqrt(101)$ , which does not simplify any further, besides to rationalize the denominator if desired, giving $\frac{- 9 \sqrt{101}}{101}$ $(-9sqrt(101))/101$ .

Hope that helps!

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What is the projection of $< 7, 1, 1 >$ $<7,1,1 >$ onto $< - 1, 6, - 8 >$ $<-1,6,-8 >$ ?

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Explanation:

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