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# If $\bold{a}$ and $\bold{b}$ are vectors such that $\|\bold{a}\| = 4$, $\|\bold{b}\| = 5$, and $\|\bold{a} + \bold{b}\| = 7$, then find $\| 0 942 4 • 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 +25646 +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 $$\text{Let \vec{a}=\dbinom{x_a}{y_a} } \\ \text{Let \vec{b}=\dbinom{x_b}{y_b} } \\ \text{Let \vec{a}\cdot \vec{b}= x_a\cdot x_b+y_a\cdot y_b } \\ \text{Let \vec{a}+\vec{b}=\dbinom{x_a+x_b}{y_a+y_b} } \\ \text{Let 2\vec{a}-3\vec{b}=\dbinom{2x_a-3x_b}{2y_a-3y_b} }$$ $$\begin{array}{|rcll|} \hline ||\vec{a}+\vec{b}|| &=& 7 \\ ||\vec{a}+\vec{b}||^2 &=& 7^2 \\ ||\vec{a}+\vec{b}||^2 &=& 49 \quad | \quad ||a+b||^2 = (x_a+x_b)^2 + (y_a+y_b)^2 \\ (x_a+x_b)^2 + (y_a+y_b)^2 &=& 49 \\ x_a^2 +2x_ax_b+x_b^2 + y_a^2 + 2y_ay_b+y_b^2 &=& 49 \\ x_a^2+y_a^2+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ 16+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ 16+25 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ 41 +2\vec{a}\cdot \vec{b} &=& 49 \\ 2\vec{a}\cdot \vec{b} &=& 8 \\ \mathbf{\vec{a}\cdot \vec{b}} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}$$ $$\begin{array}{|rcll|} \hline ||2\vec{a}-3\vec{b}|| &=& ||\dbinom{2x_a-3x_b}{2y_a-3y_b} || \\ ||2\vec{a}-3\vec{b}||^2 &=& (2x_a-3x_b)^2 + (2y_a-3y_b)^2 \\ &=& 4x_a^2 -12x_ax_b + 9 x_b^2 + 4y_a^2 - 12y_ay_b + 9 y_b^2 \\ &=& 4(x_a^2+ y_a^2) + 9( x_b^2+y_b^2) -12(x_ax_b + y_ay_b) \\\\ && x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ && x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ && x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ \\ &=& 4\cdot 16 + 9\cdot 25 -12\vec{a}\cdot \vec{b} \quad | \quad \vec{a}\cdot \vec{b} =4\\ &=& 64 + 225 -12\cdot 4 \\ &=& 289 - 48\\ &=& 241 \\ \mathbf{||2\vec{a}-3\vec{b}||} &\mathbf{=}& \mathbf{\sqrt{241 }} \\ \hline \end{array}$$ Feb 5, 2019 ### 4+0 Answers #1 +28025 0 See: https://web2.0calc.com/questions/pls-help-fast_1 Feb 5, 2019 #2 +25646 +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

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

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

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

heureka Feb 5, 2019
#3
+114090
+1

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

CPhill  Feb 5, 2019
#4
+25646
+6

Thank you, CPhill

heureka  Feb 6, 2019