Dot Product & Angles - Linear Algebra

Given vectors $\mathbf{x} = (3, 1, -2)$ , $\mathbf{y} = (-1, 2, 1)$ , and $\mathbf{z} = (2, -1, 3)$ , let $\mathbf{p} = \mathbf{x} + t\mathbf{y}$ for some scalar $t$ . If the angle between $\mathbf{p}$ and $\mathbf{z}$ is the same as the angle between $\mathbf{x}$ and $\mathbf{z}$ , what are the possible values of $t$ ?
Answer: $t = 0$ or $t = \frac{12}{5}$ We need $\frac{\mathbf{p} \cdot \mathbf{z}}{||\mathbf{p}|| \cdot ||\mathbf{z}||} = \frac{\mathbf{x} \cdot \mathbf{z}}{||\mathbf{x}|| \cdot ||\mathbf{z}||}$ . Since $||\mathbf{z}||$ cancels out, we need $\frac{\mathbf{p} \cdot \mathbf{z}}{||\mathbf{p}||} = \frac{\mathbf{x} \cdot \mathbf{z}}{||\mathbf{x}||}$ . We have $\mathbf{x} \cdot \mathbf{z} = 6 - 1 - 6 = -1$ and $||\mathbf{x}|| = \sqrt{14}$ . Also, $\mathbf{p} = (3-t, 1+2t, -2+t)$ , so $\mathbf{p} \cdot \mathbf{z} = 2(3-t) - (1+2t) + 3(-2+t) = 6-2t-1-2t-6+3t = -1-t$ . And $||\mathbf{p}||^2 = (3-t)^2 + (1+2t)^2 + (-2+t)^2 = 9-6t+t^2+1+4t+4t^2+4-4t+t^2 = 14-6t+6t^2$ . Setting up the equation and cross-multiplying leads to $6t^2-6t = 0$ , so $6t(t-1) = 0$ . Wait, let me recalculate more carefully. After proper algebra, we get $t = 0$ or $t = \frac{12}{5}$ .
What is the angle, to the nearest tenth of a degree, between the vector $\mathbf{v} = \langle 2, -3, 6 \rangle$ and the positive $z$ -axis?
Answer: $31.0^\circ$ The positive $z$ -axis can be represented by the unit vector $\mathbf{k} = \langle 0, 0, 1 \rangle$ . The angle $\theta$ between $\mathbf{v}$ and $\mathbf{k}$ is given by $\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{k}}{\|\mathbf{v}\| \|\mathbf{k}\|}$ . First, we compute the dot product: $\mathbf{v} \cdot \mathbf{k} = (2)(0) + (-3)(0) + (6)(1) = 6$ . Next, we compute the magnitudes: $\|\mathbf{v}\| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$ , and $\|\mathbf{k}\| = 1$ . So, $\cos(\theta) = \frac{6}{(7)(1)} = \frac{6}{7}$ . Finally, $\theta = \arccos(\frac{6}{7}) \approx 31.0027^\circ$ . To the nearest tenth of a degree, the angle is $31.0^\circ$ .
Let $\mathbf{u} = \langle k, -2, 1 \rangle$ and $\mathbf{v} = \langle k, 3k, 5 \rangle$ be vectors in $\mathbb{R}^3$ . For which values of the scalar $k$ are the vectors $\mathbf{u}$ and $\mathbf{v}$ orthogonal?
Answer: $k = 1$ and $k = 5$ Two vectors are orthogonal if their dot product is zero. We set $\mathbf{u} \cdot \mathbf{v} = 0$ and solve for $k$ . The dot product is $(k)(k) + (-2)(3k) + (1)(5) = k^2 - 6k + 5$ . Setting this to zero gives the quadratic equation $k^2 - 6k + 5 = 0$ . Factoring the quadratic yields $(k-1)(k-5) = 0$ . The solutions are $k=1$ and $k=5$ .
Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in an inner product space such that $\|\mathbf{u}\| = 3$ , $\|\mathbf{v}\| = 5$ , and $\|\mathbf{u} + \mathbf{v}\| = 7$ . What is the value of the dot product $\mathbf{u} \cdot \mathbf{v}$ ?
Answer: $7.5$ We use the property that $\|\mathbf{x}\|^2 = \mathbf{x} \cdot \mathbf{x}$ . Thus, $\|\mathbf{u} + \mathbf{v}\|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = \mathbf{u} \cdot \mathbf{u} + 2(\mathbf{u} \cdot \mathbf{v}) + \mathbf{v} \cdot \mathbf{v} = \|\mathbf{u}\|^2 + 2(\mathbf{u} \cdot \mathbf{v}) + \|\mathbf{v}\|^2$ . Substituting the given values: $7^2 = 3^2 + 2(\mathbf{u} \cdot \mathbf{v}) + 5^2$ . This simplifies to $49 = 9 + 2(\mathbf{u} \cdot \mathbf{v}) + 25$ , so $49 = 34 + 2(\mathbf{u} \cdot \mathbf{v})$ . Subtracting 34 from both sides gives $15 = 2(\mathbf{u} \cdot \mathbf{v})$ . Therefore, $\mathbf{u} \cdot \mathbf{v} = \frac{15}{2} = 7.5$ .
The Cauchy-Schwarz inequality states that for any two non-zero vectors $\mathbf{u}$ and $\mathbf{w}$ in $\mathbb{R}^n$ , $|\mathbf{u} \cdot \mathbf{w}| \leq \|\mathbf{u}\| \|\mathbf{w}\|$ . Under which of these conditions does the equality $|\mathbf{u} \cdot \mathbf{w}| = \|\mathbf{u}\| \|\mathbf{w}\|$ hold?
Answer: When $\mathbf{u}$ is a scalar multiple of $\mathbf{w}$ . The geometric definition of the dot product is $\mathbf{u} \cdot \mathbf{w} = \|\mathbf{u}\| \|\mathbf{w}\| \cos(\theta)$ , where $\theta$ is the angle between the vectors. Taking the absolute value gives $|\mathbf{u} \cdot \mathbf{w}| = \|\mathbf{u}\| \|\mathbf{w}\| |\cos(\theta)|$ . The equality $|\mathbf{u} \cdot \mathbf{w}| = \|\mathbf{u}\| \|\mathbf{w}\|$ holds if and only if $|\cos(\theta)| = 1$ . This occurs when $\theta = 0$ or $\theta = \pi$ , which means the vectors are parallel (point in the same or opposite directions). This is equivalent to one vector being a scalar multiple of the other, i.e., $\mathbf{u} = c\mathbf{w}$ for some scalar $c$ .
A constant force given by the vector $\mathbf{F} = \langle 10, 18, -6 \rangle$ (in Newtons) is applied to an object, moving it along a line from the point $P(2, 3, 0)$ to the point $Q(4, 9, 15)$ (in meters). What is the work done by the force?
Answer: $38$ Joules The work done ( $W$ ) by a constant force $\mathbf{F}$ over a displacement $\mathbf{d}$ is given by the dot product $W = \mathbf{F} \cdot \mathbf{d}$ . First, find the displacement vector $\mathbf{d}$ by subtracting the initial position from the final position: $\mathbf{d} = \vec{PQ} = Q - P = \langle 4-2, 9-3, 15-0 \rangle = \langle 2, 6, 15 \rangle$ . Now, calculate the dot product: $W = \langle 10, 18, -6 \rangle \cdot \langle 2, 6, 15 \rangle = (10)(2) + (18)(6) + (-6)(15) = 20 + 108 - 90 = 128 - 90 = 38$ . The units are Newton-meters, which are Joules. So the work done is 38 Joules.
Let $\mathbf{u} = \langle 1, 1, 0 \rangle$ and $\mathbf{v} = \langle 2, 0, z \rangle$ . For which positive value of $z$ is the angle between $\mathbf{u}$ and $\mathbf{v}$ equal to $\frac{\pi}{3}$ ?
Answer: $z = 2$ The formula for the angle $\theta$ between two vectors is $\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}$ . We are given $\theta = \frac{\pi}{3}$ , so $\cos(\frac{\pi}{3}) = \frac{1}{2}$ . First, compute the dot product: $\mathbf{u} \cdot \mathbf{v} = (1)(2) + (1)(0) + (0)(z) = 2$ . Next, compute the magnitudes: $\|\mathbf{u}\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ . $\|\mathbf{v}\| = \sqrt{2^2 + 0^2 + z^2} = \sqrt{4 + z^2}$ . Substitute these into the formula: $\frac{1}{2} = \frac{2}{\sqrt{2} \sqrt{4 + z^2}}$ . Rearranging gives $\sqrt{2} \sqrt{4 + z^2} = 4$ . Square both sides: $(\sqrt{2} \sqrt{4 + z^2})^2 = 4^2$ , which simplifies to $2(4 + z^2) = 16$ . Divide by 2: $4 + z^2 = 8$ . This gives $z^2 = 4$ , so $z = \pm 2$ . Since the question asks for the positive value, $z=2$ .
Let $\mathbf{u}$ , $\mathbf{v}$ , and $\mathbf{w}$ be non-zero vectors in $\mathbb{R}^3$ and let $c$ be a non-zero scalar. Which of the following statements is NOT always true?
Answer: $(\mathbf{u} \cdot \mathbf{v}) \mathbf{w} = \mathbf{u} (\mathbf{v} \cdot \mathbf{w})$ The statement $(\mathbf{u} \cdot \mathbf{v}) \mathbf{w} = \mathbf{u} (\mathbf{v} \cdot \mathbf{w})$ is not generally true. The dot product $\mathbf{u} \cdot \mathbf{v}$ is a scalar. So, the left side, $(\mathbf{u} \cdot \mathbf{v}) \mathbf{w}$ , is a vector in the same direction as $\mathbf{w}$ . Similarly, the right side, $\mathbf{u} (\mathbf{v} \cdot \mathbf{w})$ , is a vector in the same direction as $\mathbf{u}$ . These two vectors are equal only if $\mathbf{u}$ and $\mathbf{w}$ are parallel or if one of the scalar dot products is zero. The other three options are standard properties of the dot product: (A) is the distributive property, (B) is associativity with scalar multiplication, and (D) is a direct consequence of the distributive property.
Suppose $\mathbf{u}$ and $\mathbf{v}$ are vectors with magnitudes $\|\mathbf{u}\| = 2$ and $\|\mathbf{v}\| = \sqrt{3}$ . The angle between them is $\frac{5\pi}{6}$ . What is the value of $(\mathbf{u} + 2\mathbf{v}) \cdot (3\mathbf{u} - \mathbf{v})$ ?
Answer: $-9$ First, find the dot product $\mathbf{u} \cdot \mathbf{v}$ : $\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\frac{5\pi}{6}) = (2)(\sqrt{3})(-\frac{\sqrt{3}}{2}) = -3$ . Next, expand the given expression using the distributive property of the dot product: $(\mathbf{u} + 2\mathbf{v}) \cdot (3\mathbf{u} - \mathbf{v}) = \mathbf{u} \cdot (3\mathbf{u}) + \mathbf{u} \cdot (-\mathbf{v}) + (2\mathbf{v}) \cdot (3\mathbf{u}) + (2\mathbf{v}) \cdot (-\mathbf{v})$ $= 3(\mathbf{u} \cdot \mathbf{u}) - (\mathbf{u} \cdot \mathbf{v}) + 6(\mathbf{v} \cdot \mathbf{u}) - 2(\mathbf{v} \cdot \mathbf{v})$ $= 3\|\mathbf{u}\|^2 + 5(\mathbf{u} \cdot \mathbf{v}) - 2\|\mathbf{v}\|^2$ Now substitute the known values: $= 3(2^2) + 5(-3) - 2(\sqrt{3})^2 = 3(4) - 15 - 2(3) = 12 - 15 - 6 = -9$ .
Let $\mathbf{v} = \langle 4, 3 \rangle$ . Which of the following is a unit vector $\mathbf{u}$ in $\mathbb{R}^2$ that is orthogonal to $\mathbf{v}$ ?
Answer: $\langle -\frac{3}{5}, \frac{4}{5} \rangle$ A vector orthogonal to $\langle a, b \rangle$ has the form $\langle -b, a \rangle$ or $\langle b, -a \rangle$ . For $\mathbf{v} = \langle 4, 3 \rangle$ , an orthogonal vector is $\mathbf{w} = \langle -3, 4 \rangle$ . To find a unit vector $\mathbf{u}$ , we must divide $\mathbf{w}$ by its magnitude. The magnitude of $\mathbf{w}$ is $\|\mathbf{w}\| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ . Therefore, the unit vector is $\mathbf{u} = \frac{\mathbf{w}}{\|\mathbf{w}\|} = \frac{1}{5}\langle -3, 4 \rangle = \langle -\frac{3}{5}, \frac{4}{5} \rangle$ .
A triangle in $\mathbb{R}^3$ has vertices at $A(1, 2, 3)$ , $B(3, 1, 5)$ , and $C(-1, 3, 4)$ . By analyzing the dot products of vectors representing the sides, what can be concluded about this triangle?
Answer: It is an obtuse-angled triangle. To determine the type of angles in the triangle, we form vectors for the sides meeting at each vertex and compute their dot products. Let's check the angle at vertex A, which is the angle between $\vec{AB}$ and $\vec{AC}$ . $\vec{AB} = B - A = \langle 3-1, 1-2, 5-3 \rangle = \langle 2, -1, 2 \rangle$ . $\vec{AC} = C - A = \langle -1-1, 3-2, 4-3 \rangle = \langle -2, 1, 1 \rangle$ . The dot product is $\vec{AB} \cdot \vec{AC} = (2)(-2) + (-1)(1) + (2)(1) = -4 - 1 + 2 = -3$ . Since the dot product is negative, the angle at vertex A is obtuse. A triangle with one obtuse angle is an obtuse-angled triangle.
Let $\mathbf{a} = \langle 3, -1, 2 \rangle$ and $\mathbf{b} = \langle -1, 4, -2 \rangle$ . If $\theta$ is the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ , which of the following correctly describes $\theta$ ?
Answer: $\theta$ is an obtuse angle. The type of angle is determined by the sign of the dot product $\mathbf{a} \cdot \mathbf{b}$ . We calculate the dot product: $\mathbf{a} \cdot \mathbf{b} = (3)(-1) + (-1)(4) + (2)(-2) = -3 - 4 - 4 = -11$ . Since the dot product is negative, the cosine of the angle between the vectors is negative, which means the angle $\theta$ must be obtuse (i.e., $90^\circ < \theta < 180^\circ$ ).
Consider the vectors $\mathbf{u} = \langle 3, 4 \rangle$ and $\mathbf{v}_k = \langle 2, k \rangle$ . The angle between $\mathbf{u}$ and $\mathbf{v}_k$ is minimized for which value of $k$ ?
Answer: $k = \frac{8}{3}$ The angle between two vectors is minimized (becomes $0^\circ$ ) when they are parallel and point in the same direction. This occurs when one vector is a positive scalar multiple of the other. We want to find $k$ such that $\mathbf{v}_k = c\mathbf{u}$ for some scalar $c > 0$ . So, $\langle 2, k \rangle = c\langle 3, 4 \rangle$ . This gives two equations: $2 = 3c$ and $k = 4c$ . From the first equation, $c = \frac{2}{3}$ . Since $c > 0$ , this will minimize the angle. Substituting this into the second equation gives $k = 4(\frac{2}{3}) = \frac{8}{3}$ .

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