Problem 42
Question
The sides of a parallelogram are \(2 i+4 j-5 k\) and \(i+2 j\) \(+3 k\). The unit vector parallel to one of the diagonals size is (A) \(\frac{1}{7}(3 i+6 j-2 k)\) (B) \(\frac{1}{7}(3 i-6 j-2 k)\) (C) \(\frac{1}{7}(-3 i+6 j-2 k)\) (D) \(\frac{1}{7}(3 i+6 j+2 k)\)
Step-by-Step Solution
Verified Answer
The correct option is (A) \(\frac{1}{7}(3i + 6j - 2k)\).
1Step 1: Understand the Problem
We need to find a unit vector that is parallel to one of the diagonals of a parallelogram whose sides are represented by vectors \(2i + 4j - 5k\) and \(i + 2j + 3k\). A diagonal can be found by adding the vectors of the adjacent sides.
2Step 2: Find the Vector of the Diagonal
For a parallelogram, one of the diagonals is the sum of the vectors of its adjacent sides. Therefore, calculate the sum of \(2i + 4j - 5k\) and \(i + 2j + 3k\): \[(2i + 4j - 5k) + (i + 2j + 3k) = (2 + 1)i + (4 + 2)j + (-5 + 3)k = 3i + 6j - 2k.\] This gives us the vector \(3i + 6j - 2k\) for the diagonal.
3Step 3: Find the Magnitude of the Diagonal Vector
Calculate the magnitude of the vector \(3i + 6j - 2k\):\[\|3i + 6j - 2k\| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7.\]
4Step 4: Calculate the Unit Vector
To find the unit vector in the direction of \(3i + 6j - 2k\), divide the vector by its magnitude:\[\text{Unit vector} = \frac{1}{7}(3i + 6j - 2k).\]
5Step 5: Choose the Correct Option
Compare the unit vector \(\frac{1}{7}(3i + 6j - 2k)\) with the given options. The option that matches is (A) \(\frac{1}{7}(3i + 6j - 2k)\).
Key Concepts
Parallelogram DiagonalsUnit VectorVector Addition
Parallelogram Diagonals
Parallelograms are four-sided polygons with opposite sides that are equal in length and parallel. A unique feature of parallelograms is the way their diagonals intersect each other. To understand why the diagonals are relevant, remember that any parallelogram has two diagonals, each acting as a line of symmetry that divides the shape into two congruent triangles.
- The diagonal can be found by adding vectors, as each side of the parallelogram can be represented with a vector. These vectors represent both direction and magnitude, corresponding to the sides of the parallelogram.
- In the exercise, the sides of the parallelogram are given as vectors, which simplifies finding one diagonal, calculated as the sum of these vectors.
Unit Vector
A unit vector is a special type of vector in mathematics and physics that has a magnitude of exactly 1 unit, making its purpose purely directional.
- To compute a unit vector, you divide any non-zero vector by its magnitude. For example, if you have a vector \( \mathbf{v} \), the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is calculated as: \( \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \).
- This concept is critical for simplifying vector operations and ensuring consistency across various applications since it maintains only the direction of the original vector without considering its length.
- In our exercise, after determining the diagonal vector of the parallelogram, we turned it into a unit vector, focusing purely on direction, crucial for comparing it with given options.
Vector Addition
Vector addition is the operation of finding a resultant vector by summing two or more vectors. The resulting vector represents the combined effect of the initial ones.
- The process begins with aligning vectors head-to-tail or tail-to-head, producing a new vector from the starting point to the endpoint.
- Mathematically, this process is facilitated by adding the corresponding components of the vectors, such as in the form \( \mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2, a_3 + b_3) \).
- In vector algebra, particularly in exercises involving geometric shapes like parallelograms, vector addition is key to establishing crucial sub-elements, such as their diagonals.
Other exercises in this chapter
Problem 38
If \(x\) and \(y\) are two non-collinear vectors and \(A B C\) is a triangle with side lengths \(a, b, c\) satisfying \((20 a-15 b) x+(15 b-12 c) y+(12 c-20 a)(
View solution Problem 41
Let \(a=i+j\) and \(b=2 i-k\). The point of intersection of the lines \(r \times a=b \times a\) and \(r \times b=a \times b\) is (A) \(-i+j+k\) (B) \(3 i-j+k\)
View solution Problem 44
If \(a \times(b \times c)+(a \cdot b) b=(4-2 \beta-\sin \alpha) b+\left(\beta^{2}-1\right) c\) and \((c \cdot c) a=c\), while \(b\) and \(c\) are non-collinear,
View solution Problem 46
Let \(b=4 i+3 j\) and \(c\) be two vectors perpendicular to each other in \(x y\)-plane, then the vector in the same plane having projections 1 and 2 along \(b\
View solution