QuestionTrova il punto di intersezione tra il piano $4x - y + 2z - 3 = 0$ e la retta $x = -2 - 2t$ , $y = 1 + t$ , $z = -5 - t$ .

Studdy Solution

STEP 1

Assumptions1. The equation of the plane is given by $4x - y +z -3 =0$ . The equation of the line is given by the parametric equations $x = - -t$ , $y =1 + t$ , and $z = -5 - t$

STEP 2

To find the point of intersection between the plane and the line, we need to substitute the parametric equations of the line into the equation of the plane.
$4(-2 -2t) - (1 + t) +2(-5 - t) - =0$

STEP 3

implify the equation by distributing and combining like terms.
$-8 -8t -1 - t -10 -2t -3 =0$

STEP 4

Combine the terms to get a single equation in terms of $t$ .
$-22 -11t =0$

STEP 5

olve the equation for $t$ .
$t = -\frac{22}{11} = -2$

STEP 6

Now that we have the value of $t$ , we can substitute this into the parametric equations of the line to find the coordinates of the point of intersection.
$x = -2 -2(-2)$ $y =1 + (-2)$ $z = -5 - (-2)$

STEP 7

Calculate the values of $x$ , $y$ , and $z$ .
$x = -2 +4 =2$ $y =1 -2 = -1$ $z = -5 +2 = -3$ The point of intersection between the plane and the line is $(2, -1, -3)$ .

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QuestionTrova il punto di intersezione tra il piano $4x - y + 2z - 3 = 0$ e la retta $x = -2 - 2t$ , $y = 1 + t$ , $z = -5 - t$ .

Studdy Solution

STEP 1

Assumptions1. The equation of the plane is given by $4x - y +z -3 =0$ . The equation of the line is given by the parametric equations $x = - -t$ , $y =1 + t$ , and $z = -5 - t$

STEP 2

To find the point of intersection between the plane and the line, we need to substitute the parametric equations of the line into the equation of the plane.
$4(-2 -2t) - (1 + t) +2(-5 - t) - =0$

STEP 3

implify the equation by distributing and combining like terms.
$-8 -8t -1 - t -10 -2t -3 =0$

STEP 4

Combine the terms to get a single equation in terms of $t$ .
$-22 -11t =0$

STEP 5

olve the equation for $t$ .
$t = -\frac{22}{11} = -2$

STEP 6

Now that we have the value of $t$ , we can substitute this into the parametric equations of the line to find the coordinates of the point of intersection.
$x = -2 -2(-2)$ $y =1 + (-2)$ $z = -5 - (-2)$

STEP 7

Calculate the values of $x$ , $y$ , and $z$ .
$x = -2 +4 =2$ $y =1 -2 = -1$ $z = -5 +2 = -3$ The point of intersection between the plane and the line is $(2, -1, -3)$ .

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QuestionTrova il punto di intersezione tra il piano 4x−y+2z−3=04x - y + 2z - 3 = 04x−y+2z−3=0 e la retta x=−2−2tx = -2 - 2tx=−2−2t, y=1+ty = 1 + ty=1+t, z=−5−tz = -5 - tz=−5−t.

Studdy Solution

STEP 1

Assumptions1. The equation of the plane is given by 4x−y+z−3=04x - y +z -3 =04x−y+z−3=0 . The equation of the line is given by the parametric equations x=−−tx = - -tx=−−t, y=1+ty =1 + ty=1+t, and z=−5−tz = -5 - tz=−5−t

STEP 2

To find the point of intersection between the plane and the line, we need to substitute the parametric equations of the line into the equation of the plane.4(−2−2t)−(1+t)+2(−5−t)−=04(-2 -2t) - (1 + t) +2(-5 - t) - =04(−2−2t)−(1+t)+2(−5−t)−=0

STEP 3

implify the equation by distributing and combining like terms.−8−8t−1−t−10−2t−3=0-8 -8t -1 - t -10 -2t -3 =0−8−8t−1−t−10−2t−3=0

STEP 4

Combine the terms to get a single equation in terms of ttt.−22−11t=0-22 -11t =0−22−11t=0

STEP 5

olve the equation for ttt.t=−2211=−2t = -\frac{22}{11} = -2t=−1122​=−2

STEP 6

Now that we have the value of ttt, we can substitute this into the parametric equations of the line to find the coordinates of the point of intersection.x=−2−2(−2)x = -2 -2(-2)x=−2−2(−2)y=1+(−2)y =1 + (-2)y=1+(−2)z=−5−(−2)z = -5 - (-2)z=−5−(−2)

STEP 7

Calculate the values of xxx, yyy, and zzz.x=−2+4=2x = -2 +4 =2x=−2+4=2y=1−2=−1y =1 -2 = -1y=1−2=−1z=−5+2=−3z = -5 +2 = -3z=−5+2=−3The point of intersection between the plane and the line is (2,−1,−3)(2, -1, -3)(2,−1,−3).

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QuestionTrova il punto di intersezione tra il piano $4x - y + 2z - 3 = 0$ e la retta $x = -2 - 2t$ , $y = 1 + t$ , $z = -5 - t$ .

Assumptions1. The equation of the plane is given by $4x - y +z -3 =0$ . The equation of the line is given by the parametric equations $x = - -t$ , $y =1 + t$ , and $z = -5 - t$

To find the point of intersection between the plane and the line, we need to substitute the parametric equations of the line into the equation of the plane.
$4(-2 -2t) - (1 + t) +2(-5 - t) - =0$

implify the equation by distributing and combining like terms.
$-8 -8t -1 - t -10 -2t -3 =0$

Combine the terms to get a single equation in terms of $t$ .
$-22 -11t =0$

olve the equation for $t$ .
$t = -\frac{22}{11} = -2$

Now that we have the value of $t$ , we can substitute this into the parametric equations of the line to find the coordinates of the point of intersection.
$x = -2 -2(-2)$ $y =1 + (-2)$ $z = -5 - (-2)$

Calculate the values of $x$ , $y$ , and $z$ .
$x = -2 +4 =2$ $y =1 -2 = -1$ $z = -5 +2 = -3$ The point of intersection between the plane and the line is $(2, -1, -3)$ .