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Chapter 16: Problem 1508

If \((\underline{a}+\underline{b}) \cdot(\underline{a}-\underline{b})=63\) and \(|\underline{a}|=8|\underline{b}|\), then \(|\underline{a}|=\) (A) 8 (B) 64 (C) 16 4

Short Answer

Expert verified
The value of \(|\underline{a}|\) is 8.

Step by step solution

01

Expand the given expression using the properties of dot products.

We are given that \((\underline{a}+\underline{b}) \cdot(\underline{a}-\underline{b})=63\). We will expand this expression using the properties of dot products: \((\underline{a}+\underline{b}) \cdot(\underline{a}-\underline{b}) = \underline{a}\cdot\underline{a} - \underline{a}\cdot\underline{b} + \underline{a}\cdot\underline{b} - \underline{b}\cdot\underline{b}\) We know that \(\underline{a}\cdot\underline{a}=|\underline{a}|^{2}\) and \(\underline{b}\cdot\underline{b}=|\underline{b}|^{2}\). Also, since \(\underline{a}\cdot\underline{b} = \underline{b}\cdot\underline{a}\), the middle terms will cancel out. So, we have: \(|\underline{a}|^{2} - |\underline{b}|^{2} = 63\)
02

Use the given relationship between \(|\underline{a}|\) and \(|\underline{b}|\) to find a relationship between their squares.

We are given \(|\underline{a}|=8|\underline{b}|\). By squaring both sides of the equation, we find: \(|\underline{a}|^{2} = 64|\underline{b}|^{2}\)
03

Substitute the relationship between their squares from Step 2 into the expanded expression from Step 1.

Now, substitute the relationship between \(|\underline{a}|^{2}\) and \(|\underline{b}|^{2}\) that we found in Step 2 into the expanded expression from Step 1: \(64|\underline{b}|^{2} - |\underline{b}|^{2} = 63\) Simplify the expression: \(63|\underline{b}|^{2} = 63\)
04

Solve for \(|\underline{b}|^{2}\) and then find \(|\underline{b}|\).

Now, solve for \(|\underline{b}|^{2}\): \(|\underline{b}|^{2} = \frac{63}{63} = 1\) Since \(|\underline{b}|\) represents the magnitude of a vector, it must be non-negative. So, we have: \(|\underline{b}| = \sqrt{1} = 1\)
05

Find \(|\underline{a}|\) using the relationship between \(|\underline{a}|\) and \(|\underline{b}|\).

We know that \(|\underline{a}|=8|\underline{b}|\). Substitute the value of \(|\underline{b}|\) that we found in Step 4: \(|\underline{a}| = 8(1) = 8\) So, the value of \(|\underline{a}|\) is \(\boxed{8}\) (Answer Choice \textbf{(A)}).

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Most popular questions from this chapter

If angle between two vectors \(\mathrm{i}+\sqrt{3} \mathrm{j}\) and \(\sqrt{3 \mathrm{i}+\mathrm{a} j \text { is }(\pi / 3) \text { , }}\) then \(\mathrm{a}=\) (A) 0 (B) 3 (C) \(-3\) (D) none of these

If lines \([(\mathrm{x}-1) / 2]=[(\mathrm{y}-3) / 4]=\mathrm{z}\) and \([(x-4) / 3]=[(1-y) / 2]=[(z-1) / 1]\) are co-planer, then the equation of plane containing these two lines is (A) \(6 \mathrm{x}+\mathrm{y}+16 \mathrm{z}=9\) (B) \(6 \mathrm{x}+\mathrm{y}-16 \mathrm{z}=9\) (C) \(6 \mathrm{x}-\mathrm{y}-16 \mathrm{z}=9\) (D) \(6 \mathrm{x}-\mathrm{y}+16 \mathrm{z}=9\)

Line \([\\{\sqrt{(2) x-3 \sqrt{2}\\} / 1]}=[\\{2 \sqrt{2}-\sqrt{(2) \mathrm{y}\\} / 2], \mathrm{z}+1}=0\) direction cosine (A) \((1 / \sqrt{5}),-(2 \overline{/ \sqrt{5}}), 0\) (B) \(-(1 / \sqrt{5}),(1 / \sqrt{5}), 0\) (C) \((1 / \sqrt{5}), 0,(1 / \sqrt{5})\) (D) \((1 / \sqrt{5}),-(1 / \sqrt{5}), 0\)

Direction cosine of line \(\mathrm{x}=3-2 \mathrm{y}, \mathrm{z}=2 \mathrm{y}-1\) is (A) \([\\{(-2) / 3\\},(1 / 3),(2 / 3)]\) (B) \([\\{(-2) / 3\\},\\{(-1) / 3\\},(2 / 3)]\) (C) \([(2 / 3),(1 / 3),(2 / 3)]\) (D) None of these

The Equation of line passing through \((1,2,1)\) and \([(2 \mathrm{x}-1) / 3]=[(1-\mathrm{y}) / 3]=[(3 \mathrm{z}-2) / 5] \mathrm{is}\) (A) \([(2 \mathrm{x}-2) / 3]=[(2-\mathrm{y}) / 3]=[(3 \mathrm{z}-3) / 5]\) (B) \([(2 x+2) / 3]=[(2+y) / 3]=[(3 z+3) / 5]\) (C) \([(2 x-1) /(-3)]=[(1-y) /(-3)]=[(3 z-2) /(-1)]\) (D) None of these
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