The line, that is coplanar to the line $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$, is :
Solution
<p>Given two lines:</p>
<p>$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$</p>
<p>and </p>
<p>$\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$</p>
<p>These lines are coplanar if the determinant of the matrix</p>
<p>$$
\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix}
$$ = 0</p>
<p>Now let's apply this condition to the given problem. The given line is :</p>
<p>$\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$</p>
<p>So, the coordinates of any point on this line are $(-3, 1, 5)$ and the direction ratios are $(-3, 1, 5)$.</p>
<p>Now, let's calculate the determinants for each option and check which one equals zero.</p>
<p>For Option A :</p>
<p>$\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{4}$ </p>
<p>The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 4)$.</p>
<p>The determinant is :</p>
<p>$$
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{vmatrix}
$$</p>
<p>Applying the formula :</p>
<p>= $2(1 \times 4 - 5\times2) - 1((-3\times4) - (5\times-1)) + 0((-3\times2) - (1\times-1))$</p>
<p>= $2(4 - 10) - 1(-12 - (-5)) + 0(-6 - (-1))$</p>
<p>= $2(-6) - 1(-7) + 0(-5)$</p>
<p>= $-12 + 7 + 0$</p>
<p>= $-5$ </p>
<p>The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.</p>
<p>For Option B, we have :</p>
<p>$\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$</p>
<p>The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.</p>
<p>$$
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
$$</p>
<p>Applying the formula, we get :</p>
<p>= $2(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$
</p>
<p>= $2(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$</p>
<p>= $2(-5) - 1(-10) + 0(-5)$</p>
<p>= $-10 + 10 + 0$</p>
<p>= $0$</p>
So, the determinant for Option B equals zero, which confirms that the line in Option B is coplanar with the given line.
<p>For Option C :</p>
<p>$\frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ </p>
<p>The coordinates of any point on this line are $(1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.</p>
<p>The determinant is:</p>
<p>$$
\begin{vmatrix}
1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
4 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
$$</p>
<p>Applying the formula :</p>
<p>= $4(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$
</p>
<p>= $4(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$</p>
<p>= $4(-5) - 1(-10) + 0(-5)$</p>
<p>= $-20 + 10 + 0$</p>
<p>= $-10$</p>
<p>The determinant for Option C is not equal to zero, so this line is not coplanar with the given line.</p>
<p>For Option D :</p>
<p>$\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ </p>
<p>The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(1, 2, 5)$.</p>
<p>The determinant is :</p>
<p>$$
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{vmatrix}
$$</p>
<p>Applying the formula :</p>
<p>= $2(1*5 - 5*2) - 1((-3*5) - (5*1)) + 0((-3*2) - (1*1))$
</p>
<p>= $2(5 - 10) - 1(-15 - 5) + 0(-6 - 1)$</p>
<p>= $2(-5) - 1(-20) + 0(-7)$
</p>
<p>= $-10 + 20 + 0$</p>
<p>= $10$</p>
<p>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.</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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