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Paraphin [41]
3 years ago
6

Find the coordinates of the point (x,y,z) on the planez=4x+3y+1 which is closest to the origin.

Mathematics
1 answer:
Nataliya [291]3 years ago
8 0
Given plane Π : f(x,y,z) = 4x+3y-z = -1
Need to find point P on Π  that is closest to the origin O=(0,0,0).

Solution:
First step: check if O is on the plane Π : f(0,0,0)=0 ≠ -1 => O is not on Π
Next:
We know that the required point must lie on the normal vector <4,3,-1> passing through the origin, i.e. 
P=(0,0,0)+k<4,3,-1> = (4k,3k,-k)
For P to lie on plane &Pi; , it must satisfy
4(4k)+3(3k)-(-k)=-1
Solving for k
k=-1/26
=>
Point P is (4k,3k,-k) = (-4/26, -3/26, 1/26) = (-2/13, -3/26, 1/26)
because P is on the normal vector originating from the origin, and it satisfies the equation of plane &Pi;
Answer: P(-2/13, -3/26, 1/26) is the point on &Pi; closest to the origin.

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Select the correct answer. For what value of x is sin x = cos 19º, where 0°&lt; x &lt; 90°? OA. 38° ° O B. 71° OC. 26° O D. 19°​
Lerok [7]

Answer:

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Option B is correct answer.

Step-by-step explanation:

We need to find for what value of x is sin\: x = cos 19^{\circ}, where 0°< x < 90°?

Solving:

sin\: x = cos\: 19^{\circ}\\Finding \:cos\: 19^{\circ}=0.94551\\Putting\:value:\\sin\:x=0.94551\\Now\:taking\:sin\:inverse\\x=sin^{-1}(0.94551)\\x=71^{\circ}

So, we get x = 71°

Option B is correct answer.

5 0
2 years ago
IF TUPAC RAN 35 miles Away From BIGGIE SMALLS. IF BIGGIE ran 3 miles but didnt catch TUPAC how far away was TUPAC
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Step-by-step explanation:

4 0
4 years ago
Read 2 more answers
How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

3 0
3 years ago
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