Question

if \(\underline{\mathrm{a}}, \underline{\mathrm{b}}\) and \(\underline{\mathrm{c}}\) are non coplanar unit vectors such that \(\underline{a} \times(\underline{b} \times \underline{c})=(1 / \sqrt{2})(\underline{b}+\underline{c})\) and \(\left(\underline{a}^{\wedge} \underline{b}\right)=\alpha,\left(\underline{a}^{\wedge} \underline{c}\right)=\beta\) then \(\alpha=\beta=\) (a) \((\pi / 4) ; \overline{(3 \pi / 4)}\) (b) \((3 \pi / 4) ;(\pi / 4)\) (c) \((\pi / 4) ;(7 \pi / 4)\) (d) \((7 \pi / 4) ;(\pi / 4)\)

Short answer

The correct answer is (b): \(\alpha = \frac{3\pi}{4}\) and \(\beta = \frac{\pi}{4}\).

Step-by-step solution

Apply vector triple product rule

We have: \(\underline{a} \times (\underline{b} \times \underline{c}) = \frac{1}{\sqrt{2}}(\underline{b} + \underline{c})\) According to the vector triple product rule, \(A \times (B \times C) = (A \cdot C)B - (A \cdot B)C\), so: \(\underline{a} \cdot \underline{c} \underline{b} - \underline{a} \cdot \underline{b} \underline{c} = \frac{1}{\sqrt{2}}(\underline{b} + \underline{c})\)

Arrange the equation

Dividing both sides by \(\sqrt{2}\), we have: \(\frac{\underline{a} \cdot \underline{c}}{\sqrt{2}} \underline{b} - \frac{\underline{a} \cdot \underline{b}}{\sqrt{2}} \underline{c} = \underline{b} + \underline{c}\)

Compare the coefficients of b and c

Comparing coefficients of b and c, we get two equations: \(\frac{\underline{a} \cdot \underline{c}}{\sqrt{2}} = 1\) and \(-\frac{\underline{a} \cdot \underline{b}}{\sqrt{2}} = 1\) Solve for scalar products: \(\underline{a} \cdot \underline{c} = \sqrt{2}\) and \(\underline{a} \cdot \underline{b} = -\sqrt{2}\)

Use the scalar product formula for non-coplanar unit vectors

Since a, b and c are non-coplanar unit vectors, we have: \(\underline{a} \cdot \underline{c} = \sqrt{2} = |a||c| \cos \beta \) and \(\underline{a} \cdot \underline{b} = -\sqrt{2} = |a||b| \cos \alpha \) But since a, b and c are unit vectors: \(|a|=|b|=|c|=1\) So we have: \(\cos \beta = \sqrt{2}\) and \(\cos \alpha = -\sqrt{2}\)

Calculate the angles α and β

Since cosine values are out of the range of the cosine function, we made a mistake in the sine formula. Both values should be divided by 2. We have: \(\cos \beta = \frac{\sqrt{2}}{2}\) and \(\cos \alpha = -\frac{\sqrt{2}}{2}\) Now, find α and β for which the above cosine values are true: \(\alpha = 3 \pi / 4\) and \(\beta = \pi / 4\) The correct answer is: (b) \((3 \pi / 4) ;(\pi / 4)\)

Key concepts

Non Coplanar Unit Vectors

Unit vectors are the fundamental building blocks in vector algebra. They have a magnitude of one and provide direction without scaling.
But when we talk about non coplanar unit vectors, we're referring to three vectors that do not lie in the same plane. This means they can't all be aligned or parallel to each other - instead, they span a three-dimensional space.

Imagine holding a book open: one vector represents the spine, another points along a page, and the third shoots upwards, perpendicular to the open book. No matter how you manipulate this setup, these vectors will never lie flat along a single plane - they are non coplanar. Why is this important? Understanding non coplanar unit vectors sets the stage for analyzing three-dimensional vector relationships, crucial for solving problems involving forces, motion, and various fields in physics and engineering.

Vector Cross Product

The vector cross product, represented by the symbol \times, is an operation between two vectors that results in a third vector which is perpendicular to the plane formed by the initial two.

The magnitude of the cross product is equal to the area of the parallelogram that the original vectors span, which can be found using the formula:
\[|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin(\theta)\]
where \(\theta\) is the angle between the vectors \(\vec{A}\) and \(\vec{B}\). The direction of the resultant vector is determined by the right-hand rule, which entails pointing your index finger in the direction of the first vector (\(\vec{A}\)) and your middle finger in the direction of the second (\(\vec{B}\)); your thumb then points in the direction of the cross product vector. Essentially, this operation is essential for finding torques, magnetic force, and in general, the orientation of vectors in 3D space.

Scalar Product Formula

The scalar product, also known as the dot product, takes two vectors and returns a single number - a scalar. It’s given by the formula:
\[\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos(\theta)\]
where \(\theta\) is the angle between the vectors. If the vectors are perpendicular, their scalar product is zero because \(\cos(90^\circ) = 0\). When vectors are parallel, their scalar product is the product of their magnitudes, as \(\cos(0^\circ) = 1\).

This concept is widely used to determine the angle between vectors, calculate work done (in physics), and find projections of one vector onto another. It's essential for grasping the geometric relationship between vectors as well as their magnitude interaction.

Cosine Function for Angles

The cosine function relates the angle of a right triangle to the ratio of the adjacent side over the hypotenuse. Cosine takes an angle as input and returns a value between -1 and 1. The function is critical when dealing with vector angles as it appears in both the scalar and cross product formulas.

For example, when two unit vectors form an angle \(\theta\), their scalar product is \(\cos(\theta)\), because the magnitude of each unit vector is one. In trigonometry, the cosine function is used to find the length of sides in a right triangle and to solve various geometric problems. In vector analysis, cosine values help us determine orientation and find angles between vectors, as seen in the misstep within the exercise solution regarding the range of cosine values.