Let S be the line of intersection of two planes $x + y + z = 1$ and $2x + 3y-4z = 8$.
Finding Direction Cosines of a Line of Intersection Between Two Planes
The problem asks us to find the value of the expression \(43(l^2 - m^2 - n^2)\), where \((l, m, n)\) are the direction cosines of the line of intersection, denoted as S, formed by two given planes:
- Plane 1: \(x + y + z = 1\)
- Plane 2: \(2x + 3y - 4z = 8\)
To find the direction cosines \((l, m, n)\) of the line of intersection S, we first need to find the direction vector of the line. The line of intersection is perpendicular to the normal vectors of both planes.
Step 1: Identify the Normal Vectors of the Planes
The normal vector to a plane given by the equation \(Ax + By + Cz = D\) is \(\vec{n} = \langle A, B, C \rangle\). For the given planes:
- Normal vector of Plane 1 (\(\vec{n_1}\)): \(\vec{n_1} = \langle 1, 1, 1 \rangle\)
- Normal vector of Plane 2 (\(\vec{n_2}\)): \(\vec{n_2} = \langle 2, 3, -4 \rangle\)
Step 2: Calculate the Direction Vector of the Line S
The direction vector \(\vec{v}\) of the line of intersection S is perpendicular to both \(\vec{n_1}\) and \(\vec{n_2}\). We can find \(\vec{v}\) by taking the cross product of \(\vec{n_1}\) and \(\vec{n_2}\):
\(\vec{v} = \vec{n_1} \times \vec{n_2}\)
Calculating the cross product:
\( \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 2 & 3 & -4 \end{vmatrix} \) \( \vec{v} = \mathbf{i} \left( (1)(-4) - (1)(3) \right) - \mathbf{j} \left( (1)(-4) - (1)(2) \right) + \mathbf{k} \left( (1)(3) - (1)(2) \right) \) \( \vec{v} = \mathbf{i} (-4 - 3) - \mathbf{j} (-4 - 2) + \mathbf{k} (3 - 2) \) \( \vec{v} = \mathbf{i} (-7) - \mathbf{j} (-6) + \mathbf{k} (1) \) \( \vec{v} = \langle -7, 6, 1 \rangle \)So, the direction ratios of the line S are \(-7, 6, 1\).
Step 3: Determine the Direction Cosines (l, m, n)
The direction cosines \((l, m, n)\) are the components of the unit vector in the direction of \(\vec{v}\). First, we find the magnitude of \(\vec{v}\):
\( |\vec{v}| = \sqrt{(-7)^2 + 6^2 + 1^2} \) \( |\vec{v}| = \sqrt{49 + 36 + 1} \) \( |\vec{v}| = \sqrt{86} \)The direction cosines are:
- \(l = \frac{-7}{|\vec{v}|} = \frac{-7}{\sqrt{86}}\)
- \(m = \frac{6}{|\vec{v}|} = \frac{6}{\sqrt{86}}\)
- \(n = \frac{1}{|\vec{v}|} = \frac{1}{\sqrt{86}}\)
Step 4: Calculate the Value of the Expression \(43(l^2 - m^2 - n^2)\)
Now, we substitute the values of \(l, m, n\) into the expression. First, let's find \(l^2, m^2, n^2\):
- \(l^2 = \left(\frac{-7}{\sqrt{86}}\right)^2 = \frac{49}{86}\)
- \(m^2 = \left(\frac{6}{\sqrt{86}}\right)^2 = \frac{36}{86}\)
- \(n^2 = \left(\frac{1}{\sqrt{86}}\right)^2 = \frac{1}{86}\)
Now, calculate \(l^2 - m^2 - n^2\):
\( l^2 - m^2 - n^2 = \frac{49}{86} - \frac{36}{86} - \frac{1}{86} \) \( l^2 - m^2 - n^2 = \frac{49 - 36 - 1}{86} \) \( l^2 - m^2 - n^2 = \frac{12}{86} \)Finally, multiply this result by 43:
\( 43(l^2 - m^2 - n^2) = 43 \times \frac{12}{86} \)Since \(86 = 2 \times 43\), we can simplify:
\( 43 \times \frac{12}{2 \times 43} = \frac{12}{2} \) \( = 6 \)Conclusion
The value of the expression \(43(l^2 - m^2 - n^2)\) is 6.
