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  • If a and b are vectors such that \(||a||=4\)\(||b||=5\), and \(||a+b||=7\), then find \(||2a-3b||=\).

 Feb 5, 2019

Best Answer 

 #2
avatar+22188 
+9

 If $\bold{a}$ and $\bold{b}$ are vectors such that $\|\bold{a}\| = 4$, $\|\bold{b}\| = 5$, and
 $\|\bold{a} + \bold{b}\| = 7$, then find $\|2a-3b\|$

 

If a and b are vectors such that \(||a||=4\), \(||b||=5\), and \(||a+b||=7\), then find \(||2a-3b||\).

 

1.) trigonometric
\(x = ||2\vec{a}-3\vec{b}||\)

 

\(\begin{array}{|rcll|} \hline 7^2 &=& 4^2+5^2- 2*4*5*\cos(A) \\ 49 &=& 16+25- 40\cos(A) \\ 40\cos(A) &=& 16+25-49 \\ 40\cos(A) &=& -8 \\ \mathbf{\cos(A)} &\mathbf{=}& \mathbf{-\dfrac{1}{5}} \\\\ x^2 &=&(2\cdot 4)^2+(3\cdot 5)^2-2*8*15*\cos(180^{\circ}-A )\\ x^2& =&8^2+15^2-16*15*\cos(180^{\circ}-A )\\ x^2&=& 64+225 + 240*\cos(A) \\ x^2&=& 289 + 240*\left(-\dfrac{1}{5}\right) \\ x^2&=& 289 - 48 \\ x^2&=& 241 \\ \mathbf{x }&\mathbf{=}& \mathbf{\sqrt{241 }} \\ \hline \end{array} \)

 

2.) vectorial

\(\text{Let $\vec{a}=\dbinom{x_a}{y_a} $ } \\ \text{Let $\vec{b}=\dbinom{x_b}{y_b} $ } \\ \text{Let $\vec{a}\cdot \vec{b}= x_a\cdot x_b+y_a\cdot y_b $ } \\ \text{Let $\vec{a}+\vec{b}=\dbinom{x_a+x_b}{y_a+y_b} $ } \\ \text{Let $2\vec{a}-3\vec{b}=\dbinom{2x_a-3x_b}{2y_a-3y_b} $ } \)

 

\(\begin{array}{|rcll|} \hline ||\vec{a}+\vec{b}|| &=& 7 \\ ||\vec{a}+\vec{b}||^2 &=& 7^2 \\ ||\vec{a}+\vec{b}||^2 &=& 49 \quad | \quad ||a+b||^2 = (x_a+x_b)^2 + (y_a+y_b)^2 \\ (x_a+x_b)^2 + (y_a+y_b)^2 &=& 49 \\ x_a^2 +2x_ax_b+x_b^2 + y_a^2 + 2y_ay_b+y_b^2 &=& 49 \\ x_a^2+y_a^2+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ 16+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ 16+25 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ 41 +2\vec{a}\cdot \vec{b} &=& 49 \\ 2\vec{a}\cdot \vec{b} &=& 8 \\ \mathbf{\vec{a}\cdot \vec{b}} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline ||2\vec{a}-3\vec{b}|| &=& ||\dbinom{2x_a-3x_b}{2y_a-3y_b} || \\ ||2\vec{a}-3\vec{b}||^2 &=& (2x_a-3x_b)^2 + (2y_a-3y_b)^2 \\ &=& 4x_a^2 -12x_ax_b + 9 x_b^2 + 4y_a^2 - 12y_ay_b + 9 y_b^2 \\ &=& 4(x_a^2+ y_a^2) + 9( x_b^2+y_b^2) -12(x_ax_b + y_ay_b) \\\\ && x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ && x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ && x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ \\ &=& 4\cdot 16 + 9\cdot 25 -12\vec{a}\cdot \vec{b} \quad | \quad \vec{a}\cdot \vec{b} =4\\ &=& 64 + 225 -12\cdot 4 \\ &=& 289 - 48\\ &=& 241 \\ \mathbf{||2\vec{a}-3\vec{b}||} &\mathbf{=}& \mathbf{\sqrt{241 }} \\ \hline \end{array}\)

 

laugh

 Feb 5, 2019
 #1
avatar+18360 
+2

See:

 

https://web2.0calc.com/questions/pls-help-fast_1

 Feb 5, 2019
 #2
avatar+22188 
+9
Best Answer

 If $\bold{a}$ and $\bold{b}$ are vectors such that $\|\bold{a}\| = 4$, $\|\bold{b}\| = 5$, and
 $\|\bold{a} + \bold{b}\| = 7$, then find $\|2a-3b\|$

 

If a and b are vectors such that \(||a||=4\), \(||b||=5\), and \(||a+b||=7\), then find \(||2a-3b||\).

 

1.) trigonometric
\(x = ||2\vec{a}-3\vec{b}||\)

 

\(\begin{array}{|rcll|} \hline 7^2 &=& 4^2+5^2- 2*4*5*\cos(A) \\ 49 &=& 16+25- 40\cos(A) \\ 40\cos(A) &=& 16+25-49 \\ 40\cos(A) &=& -8 \\ \mathbf{\cos(A)} &\mathbf{=}& \mathbf{-\dfrac{1}{5}} \\\\ x^2 &=&(2\cdot 4)^2+(3\cdot 5)^2-2*8*15*\cos(180^{\circ}-A )\\ x^2& =&8^2+15^2-16*15*\cos(180^{\circ}-A )\\ x^2&=& 64+225 + 240*\cos(A) \\ x^2&=& 289 + 240*\left(-\dfrac{1}{5}\right) \\ x^2&=& 289 - 48 \\ x^2&=& 241 \\ \mathbf{x }&\mathbf{=}& \mathbf{\sqrt{241 }} \\ \hline \end{array} \)

 

2.) vectorial

\(\text{Let $\vec{a}=\dbinom{x_a}{y_a} $ } \\ \text{Let $\vec{b}=\dbinom{x_b}{y_b} $ } \\ \text{Let $\vec{a}\cdot \vec{b}= x_a\cdot x_b+y_a\cdot y_b $ } \\ \text{Let $\vec{a}+\vec{b}=\dbinom{x_a+x_b}{y_a+y_b} $ } \\ \text{Let $2\vec{a}-3\vec{b}=\dbinom{2x_a-3x_b}{2y_a-3y_b} $ } \)

 

\(\begin{array}{|rcll|} \hline ||\vec{a}+\vec{b}|| &=& 7 \\ ||\vec{a}+\vec{b}||^2 &=& 7^2 \\ ||\vec{a}+\vec{b}||^2 &=& 49 \quad | \quad ||a+b||^2 = (x_a+x_b)^2 + (y_a+y_b)^2 \\ (x_a+x_b)^2 + (y_a+y_b)^2 &=& 49 \\ x_a^2 +2x_ax_b+x_b^2 + y_a^2 + 2y_ay_b+y_b^2 &=& 49 \\ x_a^2+y_a^2+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ 16+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ 16+25 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ 41 +2\vec{a}\cdot \vec{b} &=& 49 \\ 2\vec{a}\cdot \vec{b} &=& 8 \\ \mathbf{\vec{a}\cdot \vec{b}} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline ||2\vec{a}-3\vec{b}|| &=& ||\dbinom{2x_a-3x_b}{2y_a-3y_b} || \\ ||2\vec{a}-3\vec{b}||^2 &=& (2x_a-3x_b)^2 + (2y_a-3y_b)^2 \\ &=& 4x_a^2 -12x_ax_b + 9 x_b^2 + 4y_a^2 - 12y_ay_b + 9 y_b^2 \\ &=& 4(x_a^2+ y_a^2) + 9( x_b^2+y_b^2) -12(x_ax_b + y_ay_b) \\\\ && x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ && x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ && x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ \\ &=& 4\cdot 16 + 9\cdot 25 -12\vec{a}\cdot \vec{b} \quad | \quad \vec{a}\cdot \vec{b} =4\\ &=& 64 + 225 -12\cdot 4 \\ &=& 289 - 48\\ &=& 241 \\ \mathbf{||2\vec{a}-3\vec{b}||} &\mathbf{=}& \mathbf{\sqrt{241 }} \\ \hline \end{array}\)

 

laugh

heureka Feb 5, 2019
 #3
avatar+100578 
+1

Thanks, heureka.....that is a nice comprehensive answer !!!

 

 

cool cool cool

CPhill  Feb 5, 2019
 #4
avatar+22188 
+6

Thank you, CPhill

 

laugh

heureka  Feb 6, 2019

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