Question 88 of 521MediumMCQ
The line, that is coplanar to the line \[\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\], is :
Solutions · 1
Official solution
Given two lines:
a1x−x1=b1y−y1=c1z−z1
and
a2x−x2=b2y−y2=c2z−z2
These lines are coplanar if the determinant of the matrix
\[
\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix}
\] = 0
Now let's apply this condition to the given problem. The given line is :
−3x+3=1y−1=5z−5
So, the coordinates of any point on this line are (−3,1,5) and the direction ratios are (−3,1,5).
Now, let's calculate the determinants for each option and check which one equals zero.
For Option A :
−1x+1=2y−2=4z−5
The coordinates of any point on this line are (−1,2,5) and the direction ratios are (−1,2,4).
The determinant is :
\[
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{vmatrix}
\]
Applying the formula :
= 2(1×4−5×2)−1((−3×4)−(5×−1))+0((−3×2)−(1×−1))
= 2(4−10)−1(−12−(−5))+0(−6−(−1))
= 2(−6)−1(−7)+0(−5)
= −12+7+0
= −5
The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.
For Option B, we have :
−1x+1=2y−2=5z−5
The coordinates of any point on this line are (−1,2,5) and the direction ratios are (−1,2,5).
\[
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
\]
Applying the formula, we get :
= 2(1∗5−5∗2)−1((−3∗5)−(5∗−1))+0((−3∗2)−(1∗−1))
= 2(5−10)−1(−15−(−5))+0(−6−(−1))
= 2(−5)−1(−10)+0(−5)
= −10+10+0
= 0
So, the determinant for Option B equals zero, which confirms that the line in Option B is coplanar with the given line.
For Option C :
−1x−1=2y−2=5z−5
The coordinates of any point on this line are (1,2,5) and the direction ratios are (−1,2,5).
The determinant is:
\[
\begin{vmatrix}
1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
4 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
\]
Applying the formula :
= 4(1∗5−5∗2)−1((−3∗5)−(5∗−1))+0((−3∗2)−(1∗−1))
= 4(5−10)−1(−15−(−5))+0(−6−(−1))
= 4(−5)−1(−10)+0(−5)
= −20+10+0
= −10
The determinant for Option C is not equal to zero, so this line is not coplanar with the given line.
For Option D :
1x+1=2y−2=5z−5
The coordinates of any point on this line are (−1,2,5) and the direction ratios are (1,2,5).
The determinant is :
\[
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{vmatrix}
\]
Applying the formula :
= 2(1∗5−5∗2)−1((−3∗5)−(5∗1))+0((−3∗2)−(1∗1))
= 2(5−10)−1(−15−5)+0(−6−1)
= 2(−5)−1(−20)+0(−7)
= −10+20+0
= 10
So the determinant for Option D is not equal to zero, which means the line in Option D is not coplanar with the given line.
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