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Question: Find the coordinates of the point where the line through the points \(\left( 3,-4,-5 \right)\) and \...

Find the coordinates of the point where the line through the points (3,−4,−5)\left( 3,-4,-5 \right) and (2,−3,1)\left( 2,-3,1 \right) crosses the plane 3x+2y+z+14=03x+2y+z+14=0.

Explanation

Solution

In order to solve this type of questions, as the line passes through two points, write the equation of the line passing through two points (3,−4,−5)(3,-4,-5) and (2,−3,1)(2,-3,1) using below equation
x−x1x1−x2=y−y1y1−y2=z−z1z1−z2\dfrac{x-{{x}_{1}}}{{{x}_{1}}-{{x}_{2}}}=\dfrac{y-{{y}_{1}}}{{{y}_{1}}-{{y}_{2}}}=\dfrac{z-{{z}_{1}}}{{{z}_{1}}-{{z}_{2}}}
After writing the equation of the line take a general point on the line, k, and put it in the equation of the plane. Solve it and then get the coordinates of that point.

Complete step-by-step solution:
As we know that the line passes through the points (3,−4,−5)\left( 3,-4,-5 \right) and (2,−3,1)\left( 2,-3,1 \right). So the general equation of line passes through two points (x1,y1,z1)({{x}_{1}},{{y}_{1}},{{z}_{1}}) and (x2,y2,z2)({{x}_{2}},{{y}_{2}},{{z}_{2}}) is as follows:
⇒x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\Rightarrow \dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}
In our case first point is (3,−4,−5)\left( 3,-4,-5 \right) so, x1=3,y1=−4,z1=−5{{x}_{1}}=3,{{y}_{1}}=-4,{{z}_{1}}=-5 and second point is (2,−3,1)\left( 2,-3,1 \right) so, x2=2,y2=−3,z2=1{{x}_{2}}=2,{{y}_{2}}=-3,{{z}_{2}}=1.
Hence, the equation of the line is:
∴x−32−3=y+4−3−(−4)=z+51−(−5)\therefore \dfrac{x-3}{2-3}=\dfrac{y+4}{-3-\left( -4 \right)}=\dfrac{z+5}{1-\left( -5 \right)}
⇒x−3−1=y+41=z+56\Rightarrow \dfrac{x-3}{-1}=\dfrac{y+4}{1}=\dfrac{z+5}{6}
Now, let us suppose a general point on the line.
⇒x−3−1=y+41=z+56=k\Rightarrow \dfrac{x-3}{-1}=\dfrac{y+4}{1}=\dfrac{z+5}{6}=k where kk is any constant.
⇒x−3−1=k⇒x=3−k.................(1)\Rightarrow \dfrac{x-3}{-1}=k\Rightarrow x=3-k.................(1)
Similarly, we get ⇒y+41=k⇒y=k−4............(2)\Rightarrow \dfrac{y+4}{1}=k\Rightarrow y=k-4............(2)
and ⇒z+56=k⇒z=6k−5..............(3)\Rightarrow \dfrac{z+5}{6}=k\Rightarrow z=6k-5..............(3)
Hence, a general point on the line is of form (3−k,k−4,6k−5)\left( 3-k,k-4,6k-5 \right).
Let (x,y,z)(x,y,z) be the coordinates of the point where the line crosses the plane 3x+2y+z+14=03x+2y+z+14=0
Putting the general point in the equation of the plane, we get
⇒3(3−k)+2(k−4)+(6k−5)+14=0\Rightarrow 3\left( 3-k \right)+2\left( k-4 \right)+\left( 6k-5 \right)+14=0
Now rearranging the terms of the above expression, we get
⇒9−3k+2k−8+6k−5+14=0\Rightarrow 9-3k+2k-8+6k-5+14=0
⇒5k+10=0\Rightarrow 5k+10=0
⇒k=−105=−2\Rightarrow k=\dfrac{-10}{5}= -2
Hence the value of kk after solving the equations is 2.
Now, putting the value of k=2k=2 in equation (1), we get
⇒x=2+3=5\Rightarrow x=2+3=5
Similarly, putting the value of k=2k=2 in equation (2), we get
⇒y=−2−4=−6\Rightarrow y=-2-4=-6
And, putting the value of k=2k=2 in equation (3), we get
⇒z=−6×2−5=−17\Rightarrow z=-6\times 2-5=-17
Hence we get the value of (x,y,z)\left( x,y,z \right) is equal to (5,−6,−17)\left( 5,-6,-17 \right).
So, (5,−6,−17)\left( 5,-6,-17 \right) is the correct answer.

Note: This is a typical question of line and a plane intersection one tricky thing is to get the intersection of the line with the plane. Students get stuck after getting the equation of the line. Hence, the crux is to choose a general point on the line and then substitute it into the equation of the plane. After solving, we get the desired value.