Let a line L$L$ passing through the point P(1, 1, 1)$P(1, 1, 1)$ be perpendicular to the lines (x - 4)/(4) = (y - 1)/(1) = (z - 1)/(1)$\frac{x - 4}{4} = \frac{y - 1}{1} = \frac{z - 1}{1}$ and (x - 17)/(1) = (y - 71)/(1) = (z)/(0)$\frac{x - 17}{1} = \frac{y - 71}{1} = \frac{z}{0}$. Let the line L$L$ intersect the yz-plane at the point Q$Q$. Another line parallel to L$L$ and passing through the point S(1, 0, -1)$S(1, 0, -1)$ intersects the yz-plane at the point R$R$. Then the square of the area of the parallelogram PQRS$PQRS$ is equal to

Related Formula

Core Logic

From the given lines:

Point on line 1: a₁ = - i + 2 j + 4 k$\vec{a_1} = -\hat{i} + 2\hat{j} + 4\hat{k}$ Direction vector: b₁ = α i - j - α k$\vec{b_1} = \alpha\hat{i} - \hat{j} - \alpha\hat{k}$

Point on line 2: a₂ = 0 i + 1 j + 1 k$\vec{a_2} = 0\hat{i} + 1\hat{j} + 1\hat{k}$ Direction vector: b₂ = α i + 2 j + 2α k$\vec{b_2} = \alpha\hat{i} + 2\hat{j} + 2\alpha\hat{k}$

Difference in points:

Step 1: Cross Product and Determinant

The numerator of the distance formula is the scalar triple product [ a₂- a₁ b₁ b₂]$[\vec{a_2}-\vec{a_1} \quad \vec{b_1} \quad \vec{b_2}]$:

Numerator = | arrayccc 1 & -1 & -3 α & -1 & -α α & 2 & 2α array |$= \left| \begin{array}{ccc} 1 & -1 & -3 \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{array} \right|$ (Note: The official solution uses a₁- a₂$\vec{a_1}-\vec{a_2}$, leading to [-1 1 3]$[-1 \quad 1 \quad 3]$ in the first row. We will follow that sign convention below).

Numerator = -1(-2α + 2α) - 1(2α² + α²) + 3(2α + α)$= -1(-2\alpha + 2\alpha) - 1(2\alpha^2 + \alpha^2) + 3(2\alpha + \alpha)$

Denominator = | b₁ × b₂| = | arrayccc i & j & k α & -1 & -α α & 2 & 2α array |$= |\vec{b_1} \times \vec{b_2}| = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{array} \right|$

Magnitude of denominator = √((-3α²)² + (3α)²) = √(9α⁴ + 9α²)$\sqrt{(-3\alpha^2)^2 + (3\alpha)^2} = \sqrt{9\alpha^4 + 9\alpha^2}$

Step 2: Equating to Distance

Divide numerator and denominator by 3α$3\alpha$ (assuming α ≠ 0$\alpha \neq 0$):

Square both sides:

Step 3: Solving the Quadratic

Factorizing the quadratic equation:

The sum of all possible values of α$\alpha$ is -7 + 1 = -6$-7 + 1 = -6$.

Pattern Recognition

When expanding scalar triple products with repeated scalar variables α$\alpha$, zero terms consistently appear via parallel vector components. Factoring α$\alpha$ out of the determinant speeds up calculation drastically.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Core Logic

Step 1: Intersection Points

Any point on L₁$L_1$ is (3λ + 1, 4λ + 2, bλ + a)$(3\lambda + 1, 4\lambda + 2, b\lambda + a)$. Since this point A$A$ must lie on line L$L$:

From 3λ = 2λ + 1 λ = 1$3\lambda = 2\lambda + 1 \implies \lambda = 1$. Thus A = (4, 6, 4)$A = (4, 6, 4)$. Also, substituting λ = 1$\lambda = 1$ into the third part:

Any point on L₂$L_2$ is (μ + 1, 4μ + 2, cμ + a)$(\mu + 1, 4\mu + 2, c\mu + a)$. Since this point B$B$ must lie on L$L$:

From μ = 2μ + 1 μ = -1$\mu = 2\mu + 1 \implies \mu = -1$. Thus B = (0, -2, 0)$B = (0, -2, 0)$. Substituting μ = -1$\mu = -1$ into the third part:

Step 2: Distance Equivalence

We are given that distance PA = PB$PA = PB$. P(1, 2, a)$P(1, 2, a)$, A(4, 6, 4)$A(4, 6, 4)$, B(0, -2, 0)$B(0, -2, 0)$.

Equating PA² = PB²$PA^2 = PB^2$:

Step 3: Finding Final Sum

From (1)$(1)$: a + b = 4 3 + b = 4 b = 1$a + b = 4 \implies 3 + b = 4 \implies b = 1$. From (2)$(2)$: a = c c = 3$a = c \implies c = 3$. We need a + b + c$a + b + c$:

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Core Logic

Find the foot of perpendicular N$N$ from P(3, 2, 1)$P(3, 2, 1)$ to the line L₁$L_1$: (x-1)/(1) = (y)/(2) = (z-1)/(1) = r$\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1} = r$. General point N(r+1, 2r, r+1)$N(r+1, 2r, r+1)$. Direction ratios of PN$PN$ are r-2, 2r-2, r$\langle r-2, 2r-2, r \rangle$. Since PN ⊥ L₁$PN \perp L_1$, dot product of direction ratios is 0: 1(r - 2) + 2(2r - 2) + 1(r) = 0 ⇒ 6r = 6 ⇒ r = 1$1(r - 2) + 2(2r - 2) + 1(r) = 0 \Rightarrow 6r = 6 \Rightarrow r = 1$. N(2, 2, 2)$N(2, 2, 2)$.

Execution

Q$Q$ is the image, so N$N$ is the midpoint of PQ$PQ$: (xQ + 3)/(2) = 2 ⇒ xQ = 1$\frac{x_Q + 3}{2} = 2 \Rightarrow x_Q = 1$ (yQ + 2)/(2) = 2 ⇒ yQ = 2$\frac{y_Q + 2}{2} = 2 \Rightarrow y_Q = 2$ (zQ + 1)/(2) = 2 ⇒ zQ = 3$\frac{z_Q + 1}{2} = 2 \Rightarrow z_Q = 3$ Q(1, 2, 3)$Q(1, 2, 3)$.

Find distance from Q$Q$ to line L₂$L_2$: (x-9)/(3) = (y-9)/(2) = (z-5)/(-2)$\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$. Let A(9, 9, 5)$A(9, 9, 5)$ be a point on L₂$L_2$. Vector AQ = -8, -7, -2$\vec{AQ} = \langle -8, -7, -2 \rangle$. Direction vector of L₂$L_2$ is m = 3, 2, -2$\vec{m} = \langle 3, 2, -2 \rangle$.

Distance d = | AQ × m|| m|$d = \frac{|\vec{AQ} \times \vec{m}|}{|\vec{m}|}$. Alternatively, use Pythagorean theorem: QM = √(AQ² - AM²)$QM = \sqrt{AQ^2 - AM^2}$. AQ² = (-8)² + (-7)² + (-2)² = 64 + 49 + 4 = 117$AQ^2 = (-8)^2 + (-7)^2 + (-2)^2 = 64 + 49 + 4 = 117$. Projection AM = | AQ · m| m| | = | -24 - 14 + 4√(9 + 4 + 4) | = 34√(17) = 2√(17)$AM = \left| \frac{\vec{AQ} \cdot \vec{m}}{|\vec{m}|} \right| = \left| \frac{-24 - 14 + 4}{\sqrt{9 + 4 + 4}} \right| = \frac{34}{\sqrt{17}} = 2\sqrt{17}$. AM² = 4 × 17 = 68$AM^2 = 4 \times 17 = 68$. QM = √(117 - 68) = √(49) = 7$QM = \sqrt{117 - 68} = \sqrt{49} = 7$.

Pattern Recognition

For line-to-point distances after an image reflection, immediately pivot to vector projection d = | PQ|² - | PQ · u|²$d = \sqrt{|\vec{PQ}|^2 - |\vec{PQ} \cdot \hat{u}|^2}$ to bypass tedious algebraic foot-finding a second time.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Related Formula

Parametric line equation: (x - x₁)/(a) = (y - y₁)/(b) = (z - z₁)/(c) = λ$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$

Core Logic

The line passing through P(43, α, β)$P(43, \alpha, \beta)$ with direction ratios 3, -1, 0$\langle 3, -1, 0 \rangle$ is: (x - 43)/(3) = (y - α)/(-1) = (z - β)/(0) = λ$\frac{x - 43}{3} = \frac{y - \alpha}{-1} = \frac{z - \beta}{0} = \lambda$ A general point on this line is P₁(43 + 3λ, α - λ, β)$P_1(43 + 3\lambda, \alpha - \lambda, \beta)$. This line intersects the given line r = 4+2μ, 0, -1+3μ$\vec{r} = \langle 4+2\mu, 0, -1+3\mu \rangle$. So, the intersection point P₁$P_1$ must satisfy the second line's coordinates.

Execution

Equating the coordinates: 43 + 3λ = 4 + 2μ ⇒ 3λ - 2μ = -39$43 + 3\lambda = 4 + 2\mu \Rightarrow 3\lambda - 2\mu = -39$ α - λ = 0 ⇒ α = λ$\alpha - \lambda = 0 \Rightarrow \alpha = \lambda$ β = -1 + 3μ ⇒ μ = (β + 1)/(3)$\beta = -1 + 3\mu \Rightarrow \mu = \frac{\beta + 1}{3}$

Also, the distance between P(43, α, β)$P(43, \alpha, \beta)$ and P₁(43+3λ, 0, β)$P_1(43+3\lambda, 0, \beta)$ is 13√(10)$13\sqrt{10}$. The vector PP₁ = 3λ, -λ, 0$\vec{PP_1} = \langle 3\lambda, -\lambda, 0 \rangle$. Distance squared: (3λ)² + (-λ)² + 0² = 10λ²$(3\lambda)^2 + (-\lambda)^2 + 0^2 = 10\lambda^2$. (13√(10))² = 10λ² ⇒ 1690 = 10λ² ⇒ λ² = 169 ⇒ λ = ± 13$(13\sqrt{10})^2 = 10\lambda^2 \Rightarrow 1690 = 10\lambda^2 \Rightarrow \lambda^2 = 169 \Rightarrow \lambda = \pm 13$. Since α = λ$\alpha = \lambda$, we have α = 13$\alpha = 13$ or α = -13$\alpha = -13$. Using λ = 13$\lambda = 13$: 3(13) - 2μ = -39 ⇒ 39 + 39 = 2μ ⇒ μ = 39$3(13) - 2\mu = -39 \Rightarrow 39 + 39 = 2\mu \Rightarrow \mu = 39$. Then β = -1 + 3(39) = 116$\beta = -1 + 3(39) = 116$ (But we need β < 0$\beta < 0$, so this is rejected).

Using λ = -13$\lambda = -13$: 3(-13) - 2μ = -39 ⇒ -39 - 2μ = -39 ⇒ μ = 0$3(-13) - 2\mu = -39 \Rightarrow -39 - 2\mu = -39 \Rightarrow \mu = 0$. Then β = -1 + 3(0) = -1$\beta = -1 + 3(0) = -1$ (This satisfies β < 0$\beta < 0$). So α = -13$\alpha = -13$ and β = -1$\beta = -1$.

Calculate α² + β²$\alpha^2 + \beta^2$: α² + β² = (-13)² + (-1)² = 169 + 1 = 170$\alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = 170$.

Pattern Recognition

When asked for distance "along a line", parameterize the directional vector from the start point directly to the target line, equate dimensions, and let the required distance dictate the scalar multiplier.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry