Vector Basics

Concept:
Three-Dimensional Geometry - Shortest Distance Between Two Lines.
The shortest distance between lines and is given by: $ 0 0 \vec{a}_1 = 2\hat{i} + 6\hat{j} + 3\hat{k} \vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k} 2\hat{j}-3\hat{k} \vec{a}_2 = 0\hat{i} + 2\hat{j} - 3\hat{k} \vec{b}_2 = 1\hat{i} + 2\hat{j} + 3\hat{k} \vec{a}_2 - \vec{a}_1 \vec{b}_1 \times \vec{b}_2 0 0 $

Concept:
Vectors - Distance between points and Perimeter.
The length of a vector joining points and is the magnitude .
Step 1: Extract the coordinates of the vertices.
Let the vertices of the triangle be , and . From the position vectors, their coordinate points are:

Step 2: Calculate the length of side AB.
Find the vector : $ \vec{BC} = \text{Position of } C - \text{Position of } B \vec{CA} = \text{Position of } A - \text{Position of } C $

Concept:
Vectors - Vector Algebra and Dot Product.
The magnitude of a sum of vectors satisfies , where is the angle between and .
Step 1: Rearrange the given vector equation.
We are given the relation .
Isolate the vector on one side to group the vectors we want to find the angle between ( and ): $ \vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos\theta |\vec{a}|=2 |\vec{c}|=3 |\vec{b}|=\sqrt{13+6\sqrt{3}} \cos\theta \theta \cos\theta = \frac{\sqrt{3}}{2} \theta [0, \pi] $

Concept:
Three-Dimensional Geometry - Equation of a Line through Two Points.
The symmetric form of a line passing through two points and is: $ A(x_1, y_1, z_1) (-1, 6, 5) B(x_2, y_2, z_2) (-2, 4, 3) (a, b, c) a = x_2 - x_1 = -2 - (-1) = -1 b = y_2 - y_1 = 4 - 6 = -2 c = z_2 - z_1 = 3 - 5 = -2 (-1, -2, -2) A(-1, 6, 5) -x -y -z -1 \frac{-(x+1)}{-1} \implies \frac{-x-1}{-1} \frac{-(y-6)}{-2} \implies \frac{6-y}{-2} \frac{-(z-5)}{-2} \implies \frac{-z+5}{-2} $ This perfectly matches Option E (and Option B, as they are printed identically).

Concept:
Vectors - Cross Product and Orthogonality.
If a vector is perpendicular to two other non-collinear vectors, it must be parallel to their cross product.
Step 1: Establish the relationship between vectors.
We are given and .
This implies that vector is perpendicular to both and .
Therefore, must be parallel to the normal of the plane containing and , which is given by their cross product .
We can write: , where is a scalar constant.
Step 2: Take the magnitude of both sides.
$ |\vec{b}\times\vec{c}| |\vec{b}\times\vec{c}| = |\vec{b}||\vec{c}|\sin\theta |\vec{b}|=3, |\vec{c}|=4 \theta = \frac{\pi}{3} \lambda |\vec{a}| = 2 |\vec{b}\times\vec{c}| = 6\sqrt{3} \lambda \vec{a} \lambda \vec{a} = \lambda (\vec{b}\times\vec{c}) $

Concept:
Three-Dimensional Geometry - Direction Cosines.
If a line has direction ratios , its direction cosines , which correspond to , are given by: $ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} (a, b, c) a = 2 b = 3 c = \sqrt{7} r = \sqrt{a^2 + b^2 + c^2} \cos\alpha, \cos\beta, \cos\gamma r \cos\alpha = \frac{a}{r} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}} \cos\beta = \frac{b}{r} = \frac{3}{2\sqrt{5}} \cos\gamma = \frac{c}{r} = \frac{\sqrt{7}}{2\sqrt{5}} \cos\alpha, \cos\beta, \cos\gamma $

Concept:
Three-Dimensional Geometry - Vector Equation of a Line.
The standard vector equation of a line is , where is a position vector and is the direction vector. The line is always parallel to its direction vector.
Step 1: Expand the given line equation.
Given: Distribute the scalar terms into their respective unit vectors: $ \hat{i}, \hat{j}, \hat{k} t t t \vec{r} = \vec{a} + t\vec{d} \vec{d} = 9\hat{i} + 3\hat{j} + 18\hat{k} \vec{d} 3 \vec{d} 3\hat{i} + \hat{j} + 6\hat{k}$.

Concept:
Vectors - Section Formula.
If point divides line segment internally in the ratio , its position vector is , where and are position vectors of and .
Step 1: Identify the position vectors and the ratio.
Let origin be . The position vectors are:

The ratio is .
Step 2: Apply the section formula to find .
$ \vec{a} \vec{b} \vec{PB} P B \vec{PB} = \vec{OB} - \vec{OP} \frac{8}{8} $

Concept:
• Equate coefficients of

Step 1: Expand RHS


Step 2: Equate components


Step 3: Solve equations
From (2) and (3): Substitute in (1):

Step 4: Find required value
Final Conclusion: