Let a line passing through the point (4,1,0)$(4,1,0)$ intersect the line L₁:(x-1)/(2)=(y-2)/(3)=(z-3)/(4)$L_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point A(α, β, γ)$A(\alpha, \beta, \gamma)$ and the line L₂:x-6=y=-z+4$L_{2}:x-6=y=-z+4$ at the point B(a, b, c)$B(a, b, c)$. Then the value of the determinant vmatrix 1 & 0 & 1 α & β & γ a & b & c vmatrix$\begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix}$ is equal to:

Related Formula

Shortest distance between perpendicular axes vectors:

Core Logic

Represent equations of straight lines symmetrically using vector directions [cite: 1418]: L₁: (x-1)/(0) = (y-2)/(0) = (z-3)/(1)$$L_1: \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$$ [cite: 1418] L₂: (x-λ)/(0) = (y-5)/(1) = (z-6)/(0)$$L_2: \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$$ [cite: 1418]

Evaluating the standard shortest distance configuration formula[cite: 1419, 1420]: S.D. = |λ - 1| = 3 λ - 1 = ± 3$$\text{S.D.} = |\lambda - 1| = 3 \implies \lambda - 1 = \pm 3$$ [cite: 1420] λ = 4 or λ = -2$$\lambda = 4 \quad \text{or} \quad \lambda = -2$$ [cite: 1420]

Given the condition λ₂ < λ₁$\lambda_2 < \lambda_1$ [cite: 698]: λ₁ = 4, λ₂ = -2$$\lambda_1 = 4, \quad \lambda_2 = -2$$ [cite: 1421, 1422]

Step 1: Distance calculation from line

We need to find the square of distance from point P(4, -2, 7)$P(4, -2, 7)$ to line L₁$L_1$ [cite: 1424]. Any general matching point coordinates tracking along path L₁$L_1$ look like Q(1, 2, t+3)$Q(1, 2, t+3)$ [cite: 1424].

Form a perpendicular projection vector condition [cite: 1425]: PQ = (-3, 4, t-4)$$\vec{PQ} = (-3, 4, t-4)$$ [cite: 1425] Since PQ · k = 0 t-4 = 0 t=4$\vec{PQ} \cdot \hat{k} = 0 \implies t-4 = 0 \implies t=4$ [cite: 1425, 1426].

Thus, the foot of perpendicular is Q(1, 2, 7)$Q(1, 2, 7)$ [cite: 1427].

Evaluate the squared distance component magnitude [cite: 1427]: PQ² = (4-1)² + (-2-2)² + (7-7)² = 3² + (-4)² + 0 = 9 + 16 = 25$$PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0 = 9 + 16 = 25$$ [cite: 1427, 1428]

Pattern Recognition

For lines parallel directly to independent Cartesian coordinate grid lines, the shortest paths are simply direct plane projections.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

The shortest distance between two lines r = a₁ + λ b₁$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and r = a₂ + μ b₂$\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by:

Core Logic

From the given line equations: Line 1 passes through a₁ = - i + 0 j + 0 k$\vec{a}_1 = -\hat{i} + 0\hat{j} + 0\hat{k}$ along vector b₁ = 3 i + 4 j + 5 k$\vec{b}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}$. Line 2 passes through a₂ = p i + 2 j + k$\vec{a}_2 = p\hat{i} + 2\hat{j} + \hat{k}$ along vector b₂ = 2 i + 3 j + 4 k$\vec{b}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

Vector difference:

Computing the cross product b₁ × b₂$\vec{b}_1 \times \vec{b}_2$:

Magnitude | b₁ × b₂| = √(1² + (-2)² + 1²) = √(6)$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.

Step 1: Applying the Shortest Distance Value

Substitute these into the distance equation:

This yields two values for p$p$:

• p - 2 = 1 p = 3$p - 2 = 1 \implies p = 3$
• p - 2 = -1 p = 1$p - 2 = -1 \implies p = 1$

Given that a, b$a, b$ are the parameters with a < b$a < b$, we assign a = 1$a = 1$ and b = 3$b = 3$.

Step 2: Computing Latus Rectum of the Ellipse

The ellipse equation is:

Since b > a$b > a$, the formula for the length of the latus rectum is:

Pattern Recognition

Be careful with coordinate geometry variables; when an ellipse satisfies b > a$b > a$, the major axis is along the y-axis, making the latus rectum equal to (2a²)/(b)$\frac{2a^2}{b}$ instead of (2b²)/(a)$\frac{2b^2}{a}$.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Conic Sections

Core Logic

First, find the point of intersection A$A$ by solving the lines. Any point on L₁$L_1$ can be written as (λ + 7, 5, -λ + 3)$(\lambda + 7, 5, -\lambda + 3)$. Substituting this point into the equation for L₂$L_2$:

Thus, the intersection point is A = (7, 5, 3)$A = (7, 5, 3)$.

Step 1: Calculating the Angle between lines

The directional vectors of lines L₁$L_1$ and L₂$L_2$ are u = i - k$\vec{u} = \hat{i} - \hat{k}$ and v = 3hati + 4hatj + 5hatk$\vec{v} = 3hat{i} + 4hat{j} + 5hat{k}$ respectively.

Now, find θ$\sin\theta$:

Step 2: Finding Area of the Triangle

The area of ABC$\triangle ABC$ given two sides and their included angle is:

Given AB = AC = √(15)$AB = AC = \sqrt{15}$:

Squaring the area:

Pattern Recognition

Since B$B$ and C$C$ lie on lines intersecting at A$A$, you don't need to determine their exact coordinates to find the area of the triangle. The standard side-angle-side area formula works perfectly using just the directional angle.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Properties of Triangles

Related Formula

Dot product of two vectors:

Condition for perpendicularity of two vectors:

Core Logic

The symmetric equation of line L$L$ is:

Any general point Q$Q$ on line L$L$ can be expressed as:

The vector from P(1, 2, 3)$P(1, 2, 3)$ to Q$Q$ is:

Since Q$Q$ is the foot of the perpendicular dropped from P$P$ onto L$L$, the vector PQ$\overrightarrow{PQ}$ is orthogonal to the line's direction vector b = 3 i + 2 j - 2 k$\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$:

Substituting λ = -1$\lambda = -1$ gives the foot of perpendicular: Q(3, 5, 9)$Q(3, 5, 9)$

Step 1: Locate Points A and B

Points A$A$ and B$B$ lie on line L$L$ at a distance d = 2√(17)$d = 2\sqrt{17}$ on either side of Q$Q$.

Let the coordinates of points on L$L$ relative to Q$Q$ be (3μ + 3, 2μ + 5, -2μ + 9)$(3\mu + 3,\, 2\mu + 5,\, -2\mu + 9)$. The distance squared from Q$Q$ is:

• For μ = 2$\mu = 2$:

• For μ = -2$\mu = -2$:

Step 2: Vector Dot Product Evaluation

Position vectors of A$A$ and B$B$ from origin O(0, 0, 0)$O(0, 0, 0)$:

Compute their scalar dot product:

Pattern Recognition

Points lying symmetrically at equal distances along a 3D line from a known central point can be found directly using parametric displacement along the unit direction vector (r = rQ ± d b$\vec{r} = \vec{r}_Q \pm d\,\hat{b}$).

Evaluation Rubric / Model Answer

Option B: 47

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry