Solveeit Logo
Login

Question

Question: The lines \(x - y + z = 1,3x - y - z + 1 = 0\) and \(x + 4y - z = 1,x + 2y + z + 1 = 0\) are: A) P...

The lines x−y+z=1,3x−y−z+1=0x - y + z = 1,3x - y - z + 1 = 0 and x+4y−z=1,x+2y+z+1=0x + 4y - z = 1,x + 2y + z + 1 = 0 are:
A) Parallel
B) Intersecting
C) Perpendicular and skew
D) Non-perpendicular and skew

Explanation

Solution

Firstly, we calculate the intersecting points of the lines. Calculate the cross product of the planes such as the cross product of plane 1 and plane 2 and the cross product of plane 3 and plane 4.

Complete step by step solution:
The given line equations are x−y+z=1,3x−y−z+1=0x - y + z = 1,3x - y - z + 1 = 0 and x+4y−z=1,x+2y+z+1=0x + 4y - z = 1,x + 2y + z + 1 = 0.
Now, we take the first two plane equations and suppose that plane 1 is p1=x−y+z=1{p_1} = x - y + z = 1 and plane 2 is p2=3x−y−z+1=0{p_2} = 3x - y - z + 1 = 0. There will be a line of intersection.
We suppose that z=0z = 0 and calculate the line of intersection of planes. Substitute the value z=0z = 0 in the expression p1=x−y+z=1{p_1} = x - y + z = 1.
x−y+(0)=1 ⇒x−y=1x - y + \left( 0 \right) = 1\\\ \Rightarrow x - y = 1 (i)
Again, we substitute the value z=0z = 0 in the expression p2=3x−y−z+1=0{p_2} = 3x - y - z + 1 = 0.
3x−y−(0)+1=0 ⇒3x−y=−13x - y - \left( 0 \right) + 1 = 0\\\ \Rightarrow 3x - y = - 1 (ii)
Now, subtract equation (i) from equation (ii):
3x−y−x+y=−1−1 ⇒2x=−2 ⇒x=−13x - y - x + y = - 1 - 1\\\ \Rightarrow 2x = - 2\\\ \Rightarrow x = - 1
We substitute the value x=−1x = - 1 in the expression x−y=1x - y = 1.
−1−y=1 ⇒−y=2 ⇒y=−2- 1 - y = 1\\\ \Rightarrow - y = 2\\\ \Rightarrow y = - 2
Hence, the line of intersection is L1=(−1,−2,0){L_1} = \left( { - 1, - 2,0} \right).
Let us suppose the direction of the line is a‾\overline a . The normal of plane 1 is n1{n_1} and normal of plane 2 is n2{n_2}.
The cross product of the normal plane is known as the direction of line.
\overline a = \overline {{n_1}} \times \overline {{n_2}} \\\ = \left| {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k}\\\ 1&{ - 1}&1\\\ 3&{ - 1}&{ - 1} \end{array}} \right|\\\ = \widehat i\left( {1 + 1} \right) - \widehat j\left( { - 1 - 3} \right) + \widehat k\left( { - 1 + 3} \right)\\\ = 2\widehat i + 4\widehat j + 2\widehat k

Hence, the equation of L1=(−1,−2,0){L_1} = \left( { - 1, - 2,0} \right) is L1=−i^−2j^+0k^+λ(2i^+4j^+2k^){L_1} = - \widehat i - 2\widehat j + 0\widehat k + \lambda \left( {2\widehat i + 4\widehat j + 2\widehat k} \right).
Take the last two plane equations and suppose plane 3 is p3=x+4y−z=1{p_3} = x + 4y - z = 1 and plane 4 is p4=x+2y+z+1=0{p_4} = x + 2y + z + 1 = 0. There will be a line of intersection.
Let us suppose z=0z = 0 and calculate the line of intersection of planes. Substitute z=0z = 0 in the expression p3=x+4y−z=1{p_3} = x + 4y - z = 1.
x+4y−0=1 ⇒x+4y=1x + 4y - 0 = 1\\\ \Rightarrow x + 4y = 1 (iii)
Again, Substitute z=0z = 0 in the expression p4=x+2y+z+1=0{p_4} = x + 2y + z + 1 = 0.
x+2y+0+1=0 ⇒x+2y=−1x + 2y + 0 + 1 = 0\\\ \Rightarrow x + 2y = - 1 (iv)
Now, subtract equation (iii) from equation (iv):
x+2y−x−4y=−1−1 ⇒−2y=−2 ⇒y=1x + 2y - x - 4y = - 1 - 1\\\ \Rightarrow - 2y = - 2\\\ \Rightarrow y = 1

Substitute y=1y = 1 in the expression x+4y=1x + 4y = 1.
⇒x+4(1)=1 ⇒x=1−4 ⇒x=−3\Rightarrow x + 4\left( 1 \right) = 1\\\ \Rightarrow x = 1 - 4\\\ \Rightarrow x = - 3
Hence, the line of intersection is L2=(−3,1,0){L_2} = \left( { - 3,1,0} \right).
Let us suppose the direction of the line is b‾\overline b . The normal of plane 3 is n3{n_3} and normal of plane 4 is n4{n_4}.
The cross product of the normal plane is known as the direction of line.
\overline b = \overline {{n_3}} \times \overline {{n_4}} \\\ = \left| {\begin{array}{*{20}{l}} {\widehat i}&{\widehat j}&{\widehat k}\\\ 1&4&{ - 1}\\\ 1&2&1 \end{array}} \right|\\\ = \widehat i\left( {4 + 2} \right) - \widehat j\left( {1 + 1} \right) + \widehat k\left( {2 - 4} \right)\\\ = 6\widehat i - 2\widehat j - 2\widehat k

Hence, the equation of L2=(−3,1,0){L_2} = \left( { - 3,1,0} \right) is L2=−3i^+j^+0k^+μ(6i^−2j^−2k^){L_2} = - 3\widehat i + \widehat j + 0\widehat k + \mu \left( {6\widehat i - 2\widehat j - 2\widehat k} \right).
Here the direction vectors a‾\overline a is not parallel to the direction vector b‾\overline b . Hence, the line is not parallel to each other.
Now, check the intersection in between the planes:
The coordinate of x in line one and line two is −1+2λ=−3+6μ- 1 + 2\lambda = - 3 + 6\mu.
Similarly, the coordinate of y in line one and line two is −2+4λ=1−2μ- 2 + 4\lambda = 1 - 2\mu and the coordinate of z in line one and line two is 2λ=−2μ2\lambda = - 2\mu which means λ=−μ\lambda = - \mu .
Substitute λ=−μ\lambda = - \mu in the expression −1+2λ=−3+6μ- 1 + 2\lambda = - 3 + 6\mu.
−1+2(−μ)=−3+6μ ⇒−1−2μ=−3+6μ ⇒8μ=−1+3 ⇒μ=14- 1 + 2\left( { - \mu } \right) = - 3 + 6\mu \\\ \Rightarrow - 1 - 2\mu = - 3 + 6\mu \\\ \Rightarrow 8\mu = - 1 + 3\\\ \Rightarrow \mu = \dfrac{1}{4}
Substitute λ=−μ\lambda = - \mu in the expression −2+4λ=1−2μ- 2 + 4\lambda = 1 - 2\mu.
−2+4(−μ)=1−2μ ⇒−2−4μ=1−2μ ⇒−2μ=1+3 ⇒μ=−2- 2 + 4\left( { - \mu } \right) = 1 - 2\mu \\\ \Rightarrow - 2 - 4\mu = 1 - 2\mu \\\ \Rightarrow - 2\mu = 1 + 3\\\ \Rightarrow \mu = - 2
Hence, the intersecting point does not exist because the value of μ\mu is different from the same value of λ\lambda . So, no real point of intersection.
Therefore, the line does not intersect with each other and is not parallel to each other; it means the lines are skew lines.
Now, check perpendicular and non-perpendicular of the direction of the line. That is:

Hence, the line is perpendicular to each other. Therefore, the option (C) is the correct option.

Note:
Here you should know what intersection points, skew points, parallel lines, and perpendicular lines. Parallel is basically the lines never intersect to each other. The perpendicular line means the lines which lie on right angles means 90∘{90^ \circ}. Skew lines are the lines which have no intersections and are not parallel.