If $A = < 7, - 3, 6 >$ $A = <7 ,-3 ,6 >$ , $B = < 4, - 7, - 5 >$ $B = <4 ,-7 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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If $A = < 7, - 3, 6 >$ $A = <7 ,-3 ,6 >$ , $B = < 4, - 7, - 5 >$ $B = <4 ,-7 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-03-24T14:49:36

$α ≅ 27, 04^{o}$ $alpha~=27,04 ^o$

Explanation:

$there are five steps:$ $"there are five steps:"$
$1 . find A-B=C$ $1. "find A-B=C"$
$2 . find dot product A.B$ $2. "find dot product A.B"$
$3 . find ||A|| (magnitude of A)$ $3. "find ||A|| (magnitude of A)"$
$4 . find ||B|| (magnitude of C)$ $4. "find ||B|| (magnitude of C)"$
$5 . use A . C = | | A | | \cdot | | C | | \cdot cos α formula$ $5. "use " A.C=||A||*||C||*cos alpha" formula"$

$S t e p - 1 :$ $Step-1:$
$A = < 7, - 3, 6 > B < 4, - 7, - 5 >$ $A= <7,-3,6>" "B<4,-7,-5>$
$C = A - B$ $C=A-B$
$C_{x} = A_{x} - B_{x} = 7 - 4; C_{x} = 3$ $C_x=A_x-B_x=7-4" ; "C_x=3$
$C - y = A_{y} - B_{y} = - 7 + 3 = - 4; C_{y} = - 4$ $C-y=A_y-B_y=-7+3=-4" ; "C_y=-4$
$C_{z} = A_{z} - B_{z} = 6 + 5 = 11; C_{z} = 11$ $C_z=A_z-B_z=6+5=11" ; "C_z=11$
$C = < 3, - 4, 11 >$ $C= <3,-4, 11>$
$Step-2:$ $"Step-2:"$
$now let's find the dot product of A.C$ $"now let's find the dot product of A.C"$
$A . C = A_{x} \cdot C_{x} + A_{y} \cdot C_{y} + A_{z} \cdot C_{z}$ $A.C=A_x*C_x+A_y*C_y+A_z*C_z$
$A . C = 7 \cdot 3 + (- 3) \cdot (- 4) + 6.11$ $A.C=7*3+(-3)*(-4)+6.11$
$A . C = 21 + 12 + 66$ $A.C=21+12+66$
$A . C = 99$ $A.C=99$
$let's find the magnitude of A and C$ $"let's find the magnitude of A and C"$
$Step-3:$ $"Step-3:"$
$| | A | | = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}} | | A | | = \sqrt{7^{2} + {(- 3)}^{2} + 6^{2}}$ $||A||=sqrt(A_x^2+A_y^2+A_z^2)" "||A||=sqrt(7^2+(-3)^2+6^2)$

$| | A | | = \sqrt{49 + 9 + 36} | | A | | = \sqrt{94}$ $||A||=sqrt(49+9+36)" "||A||=sqrt(94)$
$Step-4:$ $"Step-4:"$
$| | C | | = \sqrt{C_{x}^{2} + C_{y}^{2} + C_{z}^{2}} | | C | | = \sqrt{3^{2} + {(- 4)}^{2} + 11^{2}}$ $||C||=sqrt(C_x^2+C_y^2+C_z^2)" "||C||=sqrt(3^2+(-4)^2+11^2)$
$| | C | | = \sqrt{9 + 16 + 121} | | C | | = \sqrt{146}$ $||C||=sqrt(9+16+121)" "||C||=sqrt(146)$

$now,let's use dot product formula$ $"now,let's use dot product formula"$
$Step-5:$ $"Step-5:"$
$A . C = | | A | | \cdot | | C | | \cdot cos α$ $A.C=||A||*||C||*cos alpha$
$A . C = 99$ $A.C=99$
$| | A | | = \sqrt{94}$ $||A||=sqrt94$
$| | C | | = \sqrt{146}$ $||C||=sqrt146$

$99 = \sqrt{94} \cdot \sqrt{146} \cdot cos α$ $99=sqrt94 *sqrt 146 *cos alpha$
$99 = \sqrt{94 \cdot 146} \cdot cos α$ $99=sqrt(94*146)*cos alpha$
$99 = 111, 15 \cdot cos α$ $99=111,15*cos alpha$
$cos α = \frac{99}{111, 15} = 0, 8906882591$ $cos alpha=99/(111,15)=0,8906882591$
$α ≅ 27, 04^{o}$ $alpha~=27,04 ^o$

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If $A = < 7, - 3, 6 >$ $A = <7 ,-3 ,6 >$ , $B = < 4, - 7, - 5 >$ $B = <4 ,-7 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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If A=<7,−3,6>A = <7 ,-3 ,6 >, B=<4,−7,−5>B = <4 ,-7 ,-5 > and C=A−BC=A-B, what is the angle between A and C?

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Explanation:

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If $A = < 7, - 3, 6 >$ $A = <7 ,-3 ,6 >$ , $B = < 4, - 7, - 5 >$ $B = <4 ,-7 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?