Math Problem Statement

Найти вектор \( \vec{p} \), который ортогонален векторам \( \vec{a} = (-2, 1, -3) \) и \( \vec{b} = (-1, 3, -4) \), если \(|\vec{p}| = 2\sqrt{3}\) и вектор \( \vec{p} \) образует тупой угол с осью \(OX\).

Solution

To solve this problem, we need to find a vector p⃗\vec{p} that is orthogonal to two given vectors a⃗=(−2,1,−3)\vec{a} = (-2, 1, -3) and b⃗=(−1,3,−4)\vec{b} = (-1, 3, -4), has a magnitude of ∣p⃗∣=23|\vec{p}| = 2\sqrt{3}, and forms an obtuse angle with the OXOX axis.

Here's the solution process step-by-step:

Step 1: Find the Cross Product of a⃗\vec{a} and b⃗\vec{b}

To ensure that p⃗\vec{p} is orthogonal to both a⃗\vec{a} and b⃗\vec{b}, we can use the cross product a⃗×b⃗\vec{a} \times \vec{b}. The cross product will give us a vector orthogonal to both a⃗\vec{a} and b⃗\vec{b}.

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -3 \\ -1 & 3 & -4 \\ \end{vmatrix}$$ Calculating this determinant: 1. For $$\hat{i}$$: $$1 \cdot (-4) - (-3) \cdot 3 = -4 + 9 = 5$$ 2. For $$\hat{j}$$: $$-(-2) \cdot (-4) - (-3) \cdot (-1) = -8 - 3 = -11$$ 3. For $$\hat{k}$$: $$-2 \cdot 3 - 1 \cdot (-1) = -6 + 1 = -5$$ So, $$\vec{a} \times \vec{b} = (5, -11, -5)$$. ### Step 2: Normalize and Scale the Vector to Match the Desired Magnitude Now we have $$\vec{p} = (5, -11, -5)$$ as a candidate for the orthogonal vector. However, we need to adjust its magnitude to $$2\sqrt{3}$$. 1. Calculate the magnitude of $$\vec{a} \times \vec{b}$$: $$|\vec{a} \times \vec{b}| = \sqrt{5^2 + (-11)^2 + (-5)^2} = \sqrt{25 + 121 + 25} = \sqrt{171}$$ 2. To scale this vector to the desired magnitude of $$2\sqrt{3}$$, we use the scaling factor: $$k = \frac{2\sqrt{3}}{\sqrt{171}}$$ 3. The scaled vector $$\vec{p}$$ will then be: $$\vec{p} = k \cdot (5, -11, -5) = \frac{2\sqrt{3}}{\sqrt{171}} \cdot (5, -11, -5)$$ ### Step 3: Ensure the Vector Forms an Obtuse Angle with the $$OX$$ Axis For the vector $$\vec{p}$$ to form an obtuse angle with the $$OX$$ axis, its $$x$$-component should be negative. If the resulting $$x$$-component of $$\vec{p}$$ is positive, we simply reverse the vector to make it point in the opposite direction, thereby ensuring an obtuse angle. Would you like a detailed computation of the scaled vector $$\vec{p}$$ or assistance with other aspects of this problem? --- ### Related Questions 1. What is the geometric interpretation of the cross product? 2. How does scaling a vector affect its direction and magnitude? 3. Why does a vector form an obtuse angle with an axis when its component along that axis is negative? 4. How can we verify orthogonality using the dot product? 5. What are some applications of finding orthogonal vectors in physics and engineering? ### Tip For any vector problems involving orthogonality and specific magnitudes, using the cross product and then scaling the result is a powerful approach.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Cross Product
Orthogonality

Formulas

Cross Product: \( \vec{a} \times \vec{b} \)
Magnitude of a vector: \( |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \)

Theorems

Cross Product Orthogonality

Suitable Grade Level

Grades 10-12

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