Find the vector and parametric equations for the line through the point P=(5,−2,3) and the point Q=(2,−7,8).

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comentezq · Accepted Answer

Here think about your starting point, say it is P. Start at (5,−2,3)(5,−2,3) then to get to Q, you must travel in the direction of Q which is given by Q−P=(2,−7,8)−(5,−2,3)=(−3,−5,5). Now, traveling this direction in any scaled value gives you a line determined by the following vector equation: (5,−2,3)+t(−3,−5,5) which you can write in parametric form: x=5−3t, y=−2−3t, and z=3+5t notice that all we did was just write it in a form that represents a formula for each point (x,y,z)(x,y,z).

RizerMix · Accepted Answer

Step 1. Finding the Direction Vector (𝐯):The direction vector 𝐯 is obtained by subtracting the coordinates of P from those of Q. Hence,𝐯=[2−5−7−(−2)8−3]=[−3−55]Step 2. Parametric Equations:Let x, y, and z represent the coordinates of any point on the line. We can express the line in parametric form as:x=x0+aty=y0+btz=z0+ctwhere (x0,y0,z0) represents the coordinates of a known point on the line (in this case, P), and a, b, and c are scalar parameters.Substituting the known values, we have:x=5−3ty=−2−5tz=3+5tThus, the parametric equations of the line passing through P and Q are:x=5−3ty=−2−5tz=3+5t

Don Sumner · Accepted Answer

To find the vector and parametric equations for the line passing through point P=(5,−2,3) and point Q=(2,−7,8), we can follow these steps:1. Find the direction vector 𝐯 of the line by subtracting the coordinates of P from the coordinates of Q.𝐯=[2−5−7−(−2)8−3]=[−3−55]2. Write the vector equation of the line using the point P and the direction vector 𝐯.𝐫=[xyz]=[5−23]+t[−3−55]3. Write the parametric equations of the line by expanding the vector equation.x=5−3ty=−2−5tz=3+5tTherefore, the vector equation of the line is 𝐫=[5−23]+t[−3−55], and the parametric equations of the line are x=5−3t, y=−2−5t, and z=3+5t.

Vasquez · Accepted Answer

Result:x(t)=5−3t, y(t)=−2−5t, and z(t)=3+5tSolution:First, let&#039;s find the direction vector of the line. This can be done by subtracting the coordinates of the two points:𝐯=𝐐−𝐏=(2,−7,8)−(5,−2,3)=(−3,−5,5).The vector 𝐯 represents the direction of the line.Now, let&#039;s write the parametric equations for the line. We&#039;ll use the point P as the starting point, and we&#039;ll introduce a parameter t to represent any point on the line:𝐫(t)=𝐏+t𝐯.Substituting the values, we have:𝐫(t)=(5,−2,3)+t(−3,−5,5).Expanding this equation, we get the parametric equations:x(t)=5−3t,y(t)=−2−5t,z(t)=3+5t.Thus, the vector equation for the line is 𝐫(t)=(5,−2,3)+t(−3,−5,5), and the parametric equations for the line are x(t)=5−3t, y(t)=−2−5t, and z(t)=3+5t.

Find the vector and parametric equations for the line through the point P=(5,−2,3) and the point Q=(2,−7,8).

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