Problem 29
Question
The vectors \(a, b\) and \(c\) are equal in length and taken pairwise, they make equal angles. If \(a=i+j, b=j+\) \(k\), and \(c\) makes an obtuse angle with the base vector \(i\), then \(c\) is equal to (A) \(i+k\) (B) \(-i+4 j-k\) (C) \(\frac{-1}{3} i+\frac{4}{3} j-\frac{1}{3} k\) (D) \(\frac{1}{3} i+\frac{-4}{3} j+\frac{1}{3} k\).
Step-by-Step Solution
Verified Answer
The correct option is (C).
1Step 1: Determine Magnitude of Vectors
First, calculate the magnitude of vectors \(a\) and \(b\). Vector \(a\) is given by \(a = i + j\), so the magnitude \(|a|\) can be calculated as:\[|a| = \sqrt{1^2 + 1^2} = \sqrt{2}\]Similarly, for vector \(b = j + k\), the magnitude \(|b|\) is:\[|b| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}\]We also know \(|a| = |b| = |c|\). Thus, the magnitude of \(c\) is \(\sqrt{2}\).
2Step 2: Consider Conditions for Obtuse Angle
Since vector \(c\) makes an obtuse angle with vector \(i\), the dot product \(c \cdot i\) must be negative. The dot product is calculated as:\[c \cdot i = c_1 \times 1 + c_2 \times 0 + c_3 \times 0 = c_1\]So, \(c_1 < 0\) because the dot product must be negative for an obtuse angle.
3Step 3: Use Equivalence of Lengths and Angles
Since the vectors \(a\), \(b\), and \(c\) make equal angles pairwise, use their dot products to check.The dot product \(a \cdot b\) is:\[a \cdot b = (i + j) \cdot (j + k) = 0 \times 0 + 1 \times 1 + 0 \times 1 = 1\]For \(c\), assume a general form, but ensure \(|c| = \sqrt{2}\) and the dot products must match \(a \cdot b\).
Key Concepts
Dot ProductVector MagnitudeObtuse AnglesConcept of Equal Vectors
Dot Product
In vector algebra, the dot product, also known as the scalar product, is a mathematical operation that takes two vectors and returns a scalar. For two vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), the dot product is calculated as:\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]The dot product has several important properties:
- It is commutative, meaning \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \).
- It is distributive over vector addition, so \( \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} \).
- When two vectors are orthogonal (perpendicular), their dot product is zero.
Vector Magnitude
The magnitude of a vector, denoted as \(|\mathbf{v}|\), is a measure of its length. For a vector \( \mathbf{v} = (v_1, v_2, v_3) \), the magnitude is calculated using the square root of the sum of the squares of its components:\[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]The concept of vector magnitude is crucial because it allows us to quantify the "size" of the vector irrespective of its direction. This can be particularly useful when comparing vectors, like in our original exercise where vectors \(a\), \(b\), and \(c\) all have the same magnitude: \( \sqrt{2} \).
Magnitude is also essential for normalizing a vector (scaling it to a unit vector), which has a magnitude of 1.
Magnitude is also essential for normalizing a vector (scaling it to a unit vector), which has a magnitude of 1.
- For example, the unit vector of \(\mathbf{v}\) is given by \(\frac{\mathbf{v}}{|\mathbf{v}|}\).
Obtuse Angles
An obtuse angle is an angle that is greater than 90 degrees and less than 180 degrees. In the context of vectors, determining if two vectors form an obtuse angle involves the dot product. If the dot product \( \mathbf{a} \cdot \mathbf{b} \) is negative, then the angle between \( \mathbf{a} \) and \( \mathbf{b} \) is obtuse.
To understand why this happens, recall the equation for the dot product in terms of cosine:\[ \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta) \]If \( \cos(\theta) \) is negative, then \( \theta \) must be greater than 90 degrees because cosine values are positive for angles less than 90 degrees, zero at 90 degrees, and negative for angles greater than 90 degrees.
To understand why this happens, recall the equation for the dot product in terms of cosine:\[ \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta) \]If \( \cos(\theta) \) is negative, then \( \theta \) must be greater than 90 degrees because cosine values are positive for angles less than 90 degrees, zero at 90 degrees, and negative for angles greater than 90 degrees.
- For instance, in the problem's solution, vector \(c \) is said to make an obtuse angle with unit vector \( i \), implying \( c \cdot i < 0 \).
Concept of Equal Vectors
Equal vectors are vectors that have the same magnitude and direction. However, their position or starting point may differ. In vector notation, two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are equal if:\[ \mathbf{a} = \mathbf{b} \quad \text{implies} \quad (a_1, a_2, a_3) = (b_1, b_2, b_3) \]The concept of equal vectors is fundamental in vector algebra as it underpins the comparison of various vector quantities.
- For example, forces represented by equal vectors have the same effect irrespective of where they apply.
Other exercises in this chapter
Problem 27
A vector \(a\) is collinear with vector \(b=\left(6,-8,-7 \frac{1}{2}\right)\) and make an acute angle with the positive direction of \(z\)-axis. If \(|a|=50\),
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If the four points \(a, b, c, d\) are coplanar, then \(\left[\begin{array}{ll}b c d\end{array}\right]+\left[\begin{array}{ll}c & a & d\end{array}\right]+\left[\
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A tetrahedron has vertices at \(O(0,0,0), A(1,2,1)\), \(B(2,1,3)\) and \(C(-1,1,2)\). Then, the angle between the faces \(O A B\) and \(A B C\) will be (A) \(\c
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