Problem 10
Question
If \(a\) and \(b\) are two unit vectors, then the vector \((a+b)\) \(\times(a \times b)\) is parallel to the vector (A) \(a-b\) (B) \(a+b\) (C) \(2 a-b\) (D) \(2 a+b\)
Step-by-Step Solution
Verified Answer
The vector is parallel to option (C) \(2a - b\).
1Step 1: Understanding Unit Vectors
Unit vectors are vectors with a magnitude of 1. If \(a\) and \(b\) are unit vectors, then their dot products are simpler, i.e., \(a \cdot a = 1\) and \(b \cdot b = 1\).
2Step 2: Recall Cross Product Properties
Two important properties are: 1) The cross product of two vectors \( (c \times d) \) is a vector perpendicular to both \(c\) and \(d\). 2) The formula for triple vectors: \( (u + v) \times (u \times v) = (u \cdot v) u - (u \cdot u) v \).
3Step 3: Apply Triple Product Identity
By applying the identity, we obtain: \((a+b) \times (a \times b) = ([a \cdot b]a - [a \cdot a]b) + ([b \cdot b]a - [b \cdot a]b)\).
4Step 4: Simplify the Expanded Expression
Since \(a \cdot a = 1\) and \(b \cdot b = 1\), and \(a \cdot b\) (a scalar), simplify: \([a \cdot b]a - b + a - [a \cdot b]b \). Thus, the result is \(2a - b\).
5Step 5: Determine the Parallel Vector
The simplified result is \(2a - b\), which directly matches option (C). Therefore, the vector \((a+b) \times (a \times b)\) is parallel to \(2a - b\).
Key Concepts
Unit VectorsCross Product PropertiesTriple Product Identity
Unit Vectors
Unit vectors are essential building blocks in vector algebra. Think of them as the vectors with a special status—because their magnitude is always 1. Magnitude is just another word for length or size of the vector.
Whenever you see a unit vector like \( a \) or \( b \), you know right away that \( a \cdot a = 1 \) and \( b \cdot b = 1 \) because the dot product of a vector with itself is just the square of its magnitude. This makes calculations involving unit vectors simpler, as you often work with these perfect square magnitudes.
Whenever you see a unit vector like \( a \) or \( b \), you know right away that \( a \cdot a = 1 \) and \( b \cdot b = 1 \) because the dot product of a vector with itself is just the square of its magnitude. This makes calculations involving unit vectors simpler, as you often work with these perfect square magnitudes.
- Why they're important: Unit vectors provide direction without scaling the quantities they're used with.
- Common examples: In 3D space, the standard unit vectors are \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) representing the x, y, and z axes respectively.
Cross Product Properties
The cross product is an operation between two vectors which results in a third vector that is perpendicular to the plane of the first two. Imagine two vectors \( c \) and \( d \) lying on a flat plane like a sheet of paper. The cross product \( c \times d \) gives you a vector that sticks out of the paper!
Here are some key properties of the cross product:
Here are some key properties of the cross product:
- Perpendicularity: The result \( c \times d \) is always perpendicular to both \( c \) and \( d \).
- Magnitude: The magnitude of \( c \times d \) is equal to the area of the parallelogram that the vectors span, calculated as \( |c||d|\sin(\theta) \), where \( \theta \) is the angle between \( c \) and \( d \).
- Non-commutative: This operation is not commutative, meaning \( c \times d eq d \times c \), in fact, \( c \times d = - (d \times c) \).
Triple Product Identity
The triple product identity is a formula that simplifies the handling of the vector cross product when three vectors are involved. It's useful when you need to expand expressions like \((u + v) \times (u \times v)\).
Here's the triple product identity formula:\[(u + v) \times (u \times v) = (u \cdot v) u - (u \cdot u) v\]This powerful identity allows you to break down complex vector expressions into simpler terms that are easier to work with.
Here's the triple product identity formula:\[(u + v) \times (u \times v) = (u \cdot v) u - (u \cdot u) v\]This powerful identity allows you to break down complex vector expressions into simpler terms that are easier to work with.
- Utilization: It's most beneficial when solving vector equations as seen in problems involving torque or magnetic forces in physics, where cross products are prevalent.
- Practical Example: In the given exercise, applying the triple product identity helped transform \((a+b) \times (a \times b)\) into a simpler form to determine the parallel vector.
Other exercises in this chapter
Problem 8
If \(A B=3 i+j-k\) and \(A C=i-j+3 k .\) If the point \(P\) on the line segment \(B C\) is equidistant from \(A B\) and \(A C\), then \(A P\) is (A) \(2 i-k\) (
View solution Problem 9
\(A, B, C, D\) are four points on a plane with position vectors \(a, b, c, d\), respectively, such that \((a-d)\). \((b-c)=(b-d) \cdot(c-a)=0 .\) For \(\Delta A
View solution Problem 12
In a parallelogram \(A B C D,|A B|=a,|A D|=b\) and \(\mid A C\) \(\mid=c\). Then, \(D B \cdot A B\) has the value (A) \(\frac{3 a^{2}+b^{2}-c^{2}}{2}\) (B) \(\f
View solution Problem 14
If \(a, b, c\) are non-coplanar unit vectors such that \(a \times(b \times c)=\frac{b+c}{\sqrt{2}}\), then the angle between \(a\) and \(b\) is (A) \(\frac{3 \p
View solution