(b) Opposite sides of a square are along the lines: $$\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$$ $$\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$$ Find the area of the square if direction ratios of the other pair of opposite sides of the square are given by $<-3, 6, p>$. Also, find the value of $p$. (square, vector equation, direction ratios, area, parallel lines)

Step 1: Identify Parallel Lines and Direction Vectors

The given equations are for two lines. Since they are opposite sides of a square, they must be parallel. This means their direction vectors, $\vec{b}_{1}$ and $\vec{b}_{2}$, must be parallel. We can observe this from the given equations.

Step 2: Extract Position Vectors and Direction Vectors

From the given line equations of the form $\vec{r} = \vec{a} + t\vec{b}$, we identify the position vector of a point on each line ($\vec{a}_{1}$ and $\vec{a}_{2}$) and their common direction vector ($\vec{b}$). These lines are $L_1$ and $L_2$ respectively.

Step 3: Calculate Perpendicular Distance Between Parallel Lines

The side length of the square, let's call it $s$, is equal to the perpendicular distance $d$ between the two parallel lines. We use the formula for the distance between two parallel lines. Here, $(\vec{a}_{2} - \vec{a}_{1})$ is a vector connecting a point on $L_1$ to a point on $L_2$, and $\vec{b}$ is the common direction vector.

Step 4: Compute $(\vec{a}_{2} - \vec{a}_{1})$

First, we find the difference between the position vectors $\vec{a}_{2}$ and $\vec{a}_{1}$. This vector goes from a point on the first line to a point on the second line.

Step 5: Compute the Cross Product $(\vec{a}_{2} - \vec{a}_{1}) \times \vec{b}$

Next, we calculate the cross product of the vector $(\vec{a}_{2} - \vec{a}_{1})$ and the direction vector $\vec{b}$. The magnitude of this cross product is used in the distance formula.

Step 6: Compute Magnitude of Cross Product

We find the magnitude of the resulting vector from the cross product. This magnitude forms the numerator of our distance formula.

Step 7: Compute Magnitude of Direction Vector $\vec{b}$

We calculate the magnitude of the common direction vector $\vec{b}$. This magnitude forms the denominator of our distance formula.

Step 8: Calculate the Side Length of the Square ($s$)

Now we substitute the calculated magnitudes into the distance formula. This gives us the perpendicular distance between the two lines, which is the side length ($s$) of the square.

Step 9: Calculate the Area of the Square

The area of a square is given by the square of its side length. We square the calculated side length $s$ to find the area.

Step 10: Determine the Direction Ratios of Perpendicular Lines

The other pair of opposite sides of the square must be perpendicular to the given pair of lines. Therefore, their direction vector, which is given by $<-3, 6, p>$, must be perpendicular to the direction vector of the first pair of lines, $\vec{b} = <2, 3, 6>$. The dot product of two perpendicular vectors is zero.

Step 11: Solve for $p$

We set up the dot product equation and solve for $p$. This algebraic manipulation gives us the numerical value of $p$ which completes the direction ratios of the other pair of sides.

More from this chapter

• Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square uni
• Let L_1: (x-1)/(1) = (y-2)/(1) = (z-1)/(2) and L_2: (x+1)/(1) = (y-2)/(2) = (z)/(4) be two lines. Let L_3 be a line passing through the point (α, β,) and be per
• The line L_1 is parallel to the vector a = -3 i + 2 j + 4 k and passes through the point (7, 6, 2) and the line L_2 is parallel to the vector b = 2 i + j + 3 k
• The distance of the line (x-2)/(2)=(y-6)/(3)=(z-3)/(4) from the point (1,4,0) along the line (x)/(4)=(y-2)/(2)=(z+3)/(3) is:
• Find the shortest distance between the skew lines: r⃗₁ = (i + j + k) + λ(2i + 3j + 4k) and r⃗₂ = (2i + 3j + 5k) + μ(i + 2j + 3k).
• Find the equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes x + 2y + 3z = 4 and 2x - 3y + 4z = 5.
• Let the line passing through the points (-1, 2, 1) and parallel to the line (x-1)/(2) = (y+1)/(3) = (z)/(4) intersect the line (x+2)/(3) = (y-3)/(2) = (z-4)/(3)
• Let the shortest distance between the lines (x-3)/(3) = (y-α)/(-1) = (z-3)/(1) and (x+3)/(-3) = (y+7)/(2) = (z-β)/(4) be 3√30. Then the positive value of 5α + β

All Three Dimensional Geometry solutions →