Find parametric equations the line (Use the parameter t.) The line through the origin and the point (9, 3, −1) (x(t), y(t), z(t))=? Find the symmetric equations. a) x9=y3=z b) x+9=y+3=z−1 c) x−9=y−3=z+1 d) x9=y3=−z e) x3=y9=−z

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Find parametric equations the line (Use the parameter t.) The line through the origin and the point (9, 3, −1) (x(t), y(t), z(t))=? Find the symmetric equations. a) x9=y3=z b) x+9=y+3=z−1 c) x−9=y−3=z+1 d) x9=y3=−z e) x3=y9=−z

James Etheridge · Accepted Answer

Step 1 x=x0+at, y=y0+bt, z=z0+ctx=0+9t, y=0+3t, z=0−t For symmetric equations solve each of the parametric equations for t and set equal to each other t=x9, t=y3, t=−z x9=y3=z−1 Thus parametric equations are x=9t, y=3t, z=−t and symmetric equations are x9=y3=−z

nick1337 · Accepted Answer

Result:x9=y3=−zSolution:We can find the direction vector of the line by subtracting the coordinates of the origin from the coordinates of the point (9, 3, -1). The direction vector is given by:d→=(93−1)−(000)=(93−1)The parametric equations can be written as:x(t)=x0+dx·ty(t)=y0+dy·tz(t)=z0+dz·twhere (x0,y0,z0) is a point on the line, and (dx,dy,dz) is the direction vector.In this case, (x0,y0,z0)=(0,0,0) since the line passes through the origin. Therefore, the parametric equations become:x(t)=0+9·ty(t)=0+3·tz(t)=0+(−1)·tSimplifying the equations, we get:x(t)=9ty(t)=3tz(t)=−tThe symmetric equations can be obtained by eliminating the parameter t from the parametric equations. In this case, we can equate the ratios of x(t), y(t), and z(t) to get:x(t)9=y(t)3=z(t)−1Simplifying further, we have:x9=y3=−zTherefore, the symmetric equations for the line passing through the origin and the point (9, 3, -1) are:x9=y3=−z

Don Sumner · Accepted Answer

To find the parametric equations of the line and its symmetric equations, let&#039;s start by considering the given points: the origin, O(0, 0, 0), and the point P(9, 3, -1).The parametric equations of a line can be expressed as:x(t)=x0+aty(t)=y0+btz(t)=z0+ctwhere (x₀, y₀, z₀) is a point on the line and (a, b, c) is a direction vector of the line.To find the direction vector, we can subtract the coordinates of the two points:v→=P→−O→=(9,3,−1)−(0,0,0)=(9,3,−1)Now we can substitute the values into the parametric equations:x(t)=0+9ty(t)=0+3tz(t)&amp;amp;=0−tTherefore, the parametric equations of the line are:x(t)=9ty(t)=3tz(t)=−tTo find the symmetric equations, we can eliminate the parameter t from the parametric equations:x9=y3=z−1Multiplying each ratio by a common denominator gives:x9=y3=z−1=kwhere k is a constant.Therefore, the symmetric equations of the line are:x9=y3x9=z−1or, by eliminating fractions:3x=9y−x=9zThese are the symmetric equations of the line passing through the origin and the point (9, 3, -1).

Find parametric equations the line (Use the parameter t.)The line

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