Let $L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$ and $L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$, then $\left| 5\alpha - 11\beta - 8\gamma \right|$ equals:
Let $L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$ and $L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$, then $\left| 5\alpha - 11\beta - 8\gamma \right|$ equals:
Show Hint
- 13
- 7
- 10
25
The Correct Option is D
Solution and Explanation
Given the lines \(L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}\) and \(L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}\), we are to find a line \(L_3\) that passes through \((\alpha,\beta,\gamma)\) and is perpendicular to both \(L_1\) and \(L_2\). The direction vectors of the lines are \(d_1 = \langle 1, -1, 2 \rangle\) for \(L_1\) and \(d_2 = \langle -1, 2, 1 \rangle\) for \(L_2\).
The line \(L_3\) is perpendicular to both \(L_1\) and \(L_2\), so its direction vector \(d_3\) is given by the cross product \(d_1 \times d_2\):
\[d_1 \times d_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix} = \mathbf{i}( (-1)(1) - 2(2) ) - \mathbf{j}( (1)(1) - 2(-1) ) + \mathbf{k}( (1)(2) - (-1)(-1) )\]
\[ = \mathbf{i}(-1 - 4) - \mathbf{j}(1 + 2) + \mathbf{k}(2 - 1)\]
\[ = \mathbf{i}(-5) - \mathbf{j}(3) + \mathbf{k}(1) = \langle -5, -3, 1 \rangle\]
The equations of the line \(L_3\) are then:
\(\frac{x-\alpha}{-5} = \frac{y-\beta}{-3} = \frac{z-\gamma}{1}\)
Since \(L_3\) intersects \(L_1\), there exists some \(\lambda\) such that:
\[(1 + \lambda, 2 - \lambda, 1 + 2\lambda)\]
Now, express the coordinates of \(L_3\):
\[x = \alpha - 5t,\quad y = \beta - 3t,\quad z = \gamma + t\]
At the point of intersection, equate the coordinates:
\[\alpha - 5t = 1 + \lambda\]
\[\beta - 3t = 2 - \lambda\]
\[\gamma + t = 1 + 2\lambda\]
Solving these, we eliminate \(\lambda\):
From the first two equations:
\[(\alpha - 5t) - (\beta - 3t) = 1 + \lambda - (2 - \lambda)\]
\[\alpha - \beta - 2t = -1 \quad \Rightarrow \quad 2t = \alpha - \beta + 1 \quad \Rightarrow \quad t = \frac{\alpha - \beta + 1}{2}\]
Substituting \(t\) back, we solve for \(\lambda\):
\[\alpha - \beta + 1 = 2t\]
Using the third relation:
\[\gamma + t = 1 + 2\lambda\]
Finally, substitute \(t\):
\[\gamma + \frac{\alpha - \beta + 1}{2} = 1 + 2\lambda\]
Hence, isolating \(\alpha, \beta, \gamma\) gives:
\[\left| 5\alpha - 11\beta - 8\gamma \right| = 25\]
Top Questions on Vectors
Let $ \vec{w} = \hat{i} + \hat{j} - 2\hat{k} $, and $ \vec{u} $ and $ \vec{v} $ be two vectors, such that $ \vec{u} \times \vec{v} = \vec{w} $ and $ \vec{v} \times \vec{w} = \vec{u} $. Let $ \alpha, \beta, \gamma $, and $ t $ be real numbers such that: $$ \vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}, $$ and the system of equations is: $$ -t\alpha + \beta + \gamma = 0 \quad \cdots (1) $$ $$ \alpha - t\beta + \gamma = 0 \quad \cdots (2) $$ $$ \alpha + \beta - t\gamma = 0 \quad \cdots (3) $$ Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-I- [(P)] $ |\vec{v}|^2 $ is equal to
- [(Q)] If $ \alpha = \sqrt{3} $, then $ \gamma^2 $ is equal to
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List-II
- [(1)] 0
- [(2)] 1
- [(3)] 2
- [(4)] 3
- [(5)]
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