Key Concepts and Formulas

1. Vector Equation of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{d}$ is given by $\vec{r} = \vec{a} + \lambda \vec{d}$, where $\lambda$ is a scalar parameter. The vector $\vec{d}$ is the direction vector of the line.
2. Perpendicularity of Lines: Two lines are perpendicular if and only if their direction vectors are orthogonal. If $\vec{d_1}$ and $\vec{d_2}$ are the direction vectors of two lines, then their dot product is zero: $\vec{d_1} \cdot \vec{d_2} = 0$.
3. Vector Perpendicular to Two Vectors: A vector that is perpendicular to two given non-parallel vectors $\vec{a}$ and $\vec{b}$ is parallel to their cross product, $\vec{a} \times \vec{b}$. This property is crucial for finding the direction of a line perpendicular to two other lines.
4. Line Passing Through the Origin: A line whose equation is of the form $\vec{r} = \delta \vec{d}$ passes through the origin $(0,0,0)$ (when $\delta=0$) and has $\vec{d}$ as its direction vector. Any point on such a line will be a scalar multiple of its direction vector.

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Step-by-Step Solution

Step 1: Identify the Direction Vectors of the Given Lines

• What and Why: We extract the direction vectors from the given vector equations of the lines. This is the first step because the conditions of perpendicularity (and thus the core of the problem) relate directly to these direction vectors.

For line $L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k})$, the direction vector is: $\vec{d_1} = \hat{i}-\hat{j}+2 \hat{k}$ For line $L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k})$, the direction vector is: $\vec{d_2} = 3 \hat{i}+\hat{j}+p \hat{k}$ For line $L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k})$, the direction vector is: $\vec{d_3} = \ell \hat{i}+m \hat{j}+n \hat{k}$

• Reasoning: The coefficient of the scalar parameter ($\lambda, \mu, \delta$) in each line's equation represents its direction vector. We also note that $L_3$ passes through the origin $(0,0,0)$ because its position vector component is zero.

Step 2: Determine the Value of $p$ using the Perpendicularity of $L_1$ and $L_2$

• What and Why: We are given that $L_1$ is perpendicular to $L_2$. This condition allows us to set up an equation using the dot product of their direction vectors, which will enable us to solve for the unknown parameter $p$.

The dot product of their direction vectors must be zero: $\vec{d_1} \cdot \vec{d_2} = 0$ $(\hat{i}-\hat{j}+2 \hat{k}) \cdot (3 \hat{i}+\hat{j}+p \hat{k}) = 0$

• Math: Recall that for $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, their dot product is $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$. $(1)(3) + (-1)(1) + (2)(p) = 0$ $3 - 1 + 2p = 0$ $2 + 2p = 0$ $2p = -2$ $p = -1$
• Reasoning: By applying the perpendicularity condition, we've successfully found the value of $p$. Now we have the complete direction vector for $L_2$: $\vec{d_2} = 3 \hat{i}+\hat{j}-\hat{k}$

Step 3: Determine the Direction Vector of $L_3$

• What and Why: We are given that $L_3$ is perpendicular to both $L_1$ and $L_2$. This means its direction vector, $\vec{d_3}$, must be orthogonal to both $\vec{d_1}$ and $\vec{d_2}$. A vector that is perpendicular to two non-parallel vectors is parallel to their cross product. Calculating the cross product of $\vec{d_1}$ and $\vec{d_2}$ will give us the direction of $L_3$.

The direction vector of $L_3$, $\vec{d_3}$, must be parallel to $\vec{d_1} \times \vec{d_2}$. We can choose $\vec{d_3}$ to be equal to $\vec{d_1} \times \vec{d_2}$ (or any non-zero scalar multiple of it) for simplicity.

• Math: $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{vmatrix}$ Recall the formula for a $3 \times 3$ determinant: $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei-fh) - b(di-fg) + c(dh-eg)$. $= \hat{i}((-1)(-1) - (2)(1)) - \hat{j}((1)(-1) - (2)(3)) + \hat{k}((1)(1) - (-1)(3))$ $= \hat{i}(1 - 2) - \hat{j}(-1 - 6) + \hat{k}(1 + 3)$ $= \hat{i}(-1) - \hat{j}(-7) + \hat{k}(4)$ $= -\hat{i} + 7\hat{j} + 4\hat{k}$
• Reasoning: So, the direction vector of $L_3$, $(\ell \hat{i}+m \hat{j}+n \hat{k})$, is parallel to $(-\hat{i} + 7\hat{j} + 4\hat{k})$. This means the coordinates $(\ell, m, n)$ are proportional to $(-1, 7, 4)$.

Step 4: Find a Point that Lies on $L_3$

• What and Why: The equation of line $L_3$ is given as $\vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k})$. This form explicitly indicates that $L_3$ passes through the origin $(0,0,0)$ and its direction vector is $(\ell \hat{i}+m \hat{j}+n \hat{k})$. Therefore, any point on $L_3$ must be a scalar multiple of its direction vector. We will check the given options to find which one fits this criteria.

Since the direction vector of $L_3$ is parallel to $(-\hat{i} + 7\hat{j} + 4\hat{k})$, any point on $L_3$ can be represented as $k(-\hat{i} + 7\hat{j} + 4\hat{k})$ for some scalar $k \in \mathbb{R}$. In coordinate form, a point on $L_3$ is of the form $(-k, 7k, 4k)$.

• Checking Options:

  • (A) $(1, 7, -4)$: For this point to be on $L_3$, we would need $1 = -k$, $7 = 7k$, and $-4 = 4k$. From $1 = -k$, we get $k = -1$. Substituting $k = -1$ into $7 = 7k$ gives $7 = 7(-1) = -7$, which is false. Thus, option (A) is not a point on $L_3$.
  • (B) $(1, -7, 4)$: For this point to be on $L_3$, we would need $1 = -k$, $-7 = 7k$, and $4 = 4k$. From $1 = -k$, we get $k = -1$. Substituting $k = -1$ into $4 = 4k$ gives $4 = 4(-1) = -4$, which is false. Thus, option (B) is not a point on $L_3$.
  • (C) $(-1, 7, 4)$: For this point to be on $L_3$, we would need $-1 = -k$, $7 = 7k$, and $4 = 4k$. From $-1 = -k$, we get $k = 1$. Substituting $k = 1$ into $7 = 7k$ gives $7 = 7(1) = 7$, which is true. Substituting $k = 1$ into $4 = 4k$ gives $4 = 4(1) = 4$, which is true. Since all conditions are satisfied for $k=1$, the point $(-1, 7, 4)$ lies on $L_3$.
  • (D) $(- , 1-7,4)$: This option is malformed. Even if interpreted as $(-1, -7, 4)$, it would not satisfy the conditions (as $7k$ would be $7$, not $-7$).

• Reasoning: Only option (C) provides a consistent scalar value $k=1$ across all components when compared to the derived form of points on $L_3$. Therefore, $(-1, 7, 4)$ is the point that lies on $L_3$.

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Common Mistakes & Tips

• Cross Product Accuracy: Be extremely careful when calculating the cross product. A single sign error or incorrect determinant expansion can lead to a completely wrong direction vector.
• Scalar Multiples: Remember that a line's direction vector is unique only up to a non-zero scalar multiple. When checking options, ensure the point is a consistent scalar multiple of your derived direction vector across all coordinates.
• Line Through Origin: The form $\vec{r} = \delta \vec{d}$ is specific: it means the line passes through the origin. This simplifies finding points on the line, as any point on it must be a direct scalar multiple of its direction vector.

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Summary

We systematically determined the unknown parameter $p$ in $L_2$'s direction vector by utilizing the perpendicularity condition between $L_1$ and $L_2$. Subsequently, we found the direction vector of $L_3$ by calculating the cross product of the direction vectors of $L_1$ and $L_2$, as $L_3$ is perpendicular to both. Finally, understanding that $L_3$ passes through the origin, we identified the option that represents a point lying on $L_3$ by verifying if it is a scalar multiple of $L_3$'s direction vector. The point $(-1,7,4)$ satisfies all the given conditions.

The final answer is $\boxed{\text{(C)}}$.