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Chapter 17: Problem 1628

if \(|\underline{\mathrm{A}}|=3,|\underline{\mathrm{B}}|=4,|\underline{\mathrm{C}}|=5, \underline{\mathrm{A}} \perp(\underline{\mathrm{B}}+\underline{\mathrm{C}}), \underline{\mathrm{B}} \perp(\underline{\mathrm{C}}+\underline{\mathrm{A}})\), \(\underline{\mathrm{C}} \perp(\underline{\mathrm{A}}+\underline{\mathrm{B}})\) then \(|\underline{\mathrm{A}}+\underline{\mathrm{B}}+\underline{\mathrm{C}}|=\) (a) \(5 \sqrt{2}\) (b) \(7 \sqrt{2}\) (c) \(\sqrt{2}\) (d) \(3 \sqrt{2}\)

Short Answer

Expert verified
The magnitude of the vector \(\underline{\mathrm{A}}+\underline{\mathrm{B}}+\underline{\mathrm{C}}\) is \(5 \sqrt{2}\).

Step by step solution

01

Write down the given information

We are given: - \(|\underline{\mathrm{A}}|=3\) - \(|\underline{\mathrm{B}}|=4\) - \(|\underline{\mathrm{C}}|=5\) - \(\underline{\mathrm{A}} \perp(\underline{\mathrm{B}}+\underline{\mathrm{C}})\) - \(\underline{\mathrm{B}} \perp(\underline{\mathrm{C}}+\underline{\mathrm{A}})\) - \(\underline{\mathrm{C}} \perp(\underline{\mathrm{A}}+\underline{\mathrm{B}})\)
02

Use the properties of the dot product to find relationships between the magnitudes of the vectors

Since each of the given vectors is perpendicular to the sum of the other two, we can write the following equations using the property that the dot product of two perpendicular vectors is equal to zero: \(\underline{\mathrm{A}} \cdot (\underline{\mathrm{B}}+\underline{\mathrm{C}}) = 0\) \(\underline{\mathrm{B}} \cdot (\underline{\mathrm{C}}+\underline{\mathrm{A}}) = 0\) \(\underline{\mathrm{C}} \cdot (\underline{\mathrm{A}}+\underline{\mathrm{B}}) = 0\)
03

Expand the dot products and apply the cosine formula

Expand each of the dot products and apply the property that the cosine of the angle between two vectors equals the dot product divided by the product of the magnitudes: \(3^2 (\cos{(\angle ABC)} + \cos{(\angle BCA)}) = 0\) \(4^2 (\cos{(\angle BCA)} + \cos{(\angle CAB)}) = 0\) \(5^2 (\cos{(\angle CAB)} + \cos{(\angle ABC)}) = 0\)
04

Solve the system of equations for the cosines of the angles

Solve the system of equations for the cosines of the angles, then substitute back into the equations: \(\cos{(\angle ABC)} = -\frac{5^2}{3^2}\cos{(\angle BCA)}\) \(\cos{(\angle BCA)} = -\frac{3^2}{4^2}\cos{(\angle CAB)}\) \(\cos{(\angle CAB)} = -\frac{4^2}{5^2}\cos{(\angle ABC)}\)
05

Calculate the square of the magnitude of the sum of the vectors

Now we need to calculate the square of the magnitude of the sum of the vectors, using the property \((\underline{\mathrm{A}}+\underline{\mathrm{B}}+\underline{\mathrm{C}})^2 = |\underline{\mathrm{A}}|^2 + |\underline{\mathrm{B}}|^2 + |\underline{\mathrm{C}}|^2 + 2(\underline{\mathrm{A}}\cdot \underline{\mathrm{B}} + \underline{\mathrm{A}}\cdot \underline{\mathrm{C}} + \underline{\mathrm{B}}\cdot \underline{\mathrm{C}})\): \((\underline{\mathrm{A}}+\underline{\mathrm{B}}+\underline{\mathrm{C}})^2=3^2+4^2+5^2+2(3^2\cos{(\angle ABC)}+4^2\cos{(\angle BCA)}+5^2\cos{(\angle CAB)})\)
06

Substitute the cosines obtained in Step 4 into the expression and simplify

Substitute the cosines obtained in Step 4 into the expression: \((\underline{\mathrm{A}}+\underline{\mathrm{B}}+\underline{\mathrm{C}})^2 = 9+16+25+2(-25+9-16) = 50\)
07

Find the magnitude of the sum of the vectors

Take the square root of the result obtained in Step 6, to find the magnitude of the sum of the vectors: \(|\underline{\mathrm{A}}+\underline{\mathrm{B}}+\underline{\mathrm{C}}|=\sqrt{50}=5\sqrt{2}\) The answer is: (a) \(5 \sqrt{2}\)

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Most popular questions from this chapter

if \(\underline{a}\) and \(\underline{b}\) are vector and \(\underline{a} \cdot \underline{b}<0,|\underline{a} \cdot \underline{b}|=|\underline{a} \times \underline{b}|\) then angle between \(\underline{a}\) and \(\underline{b}\) is = (a) \(\pi\) (b) \((7 \pi / 4)\) (c) \((\pi / 4)\) (d) \((3 \pi / 4)\)

a and \(\underline{b}\) are unit vector the angle between the vectors is \(\theta\). \(|\underline{\mathrm{a}}+\underline{\mathrm{b}}|>1\). Then (a) \(\theta=(\pi / 2)\) (b) \(\theta<(\pi / 3)\) (c) \(\theta>(2 \pi / 3)\) (d) \((\pi / 2)<\theta<(2 \pi / 3)\)

a force \(\underline{F}=(2,1,-1)\) act on a practical and displaces it from the point \(\mathrm{A}(2,-1,0)\) to the point \(\mathrm{B}(2,1,0)\) then work done by force is equal to (a) 2 (b) 4 (c) 6 (d) 5

if \(2 \mathrm{i}+\mathrm{j}, \underline{\mathrm{i}}-3 \mathrm{j}, 3 \mathrm{i}+2 \mathrm{j} \& \underline{\mathrm{i}}+\lambda \mathrm{j}\) are position vectors of points \(\mathrm{A}\), \(\mathrm{B}, \mathrm{C}, \mathrm{D}\) respectively and \(\mathrm{AB} \| \mathrm{CD}\). Then \(\lambda\) is (a) 6 (b) 8 (c) \(-8\) (d) -6

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