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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

The locus of point of the plane passing through \((\alpha, \beta, \gamma)\) and intersect the axis in \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and the plane which is parallel to such plane is (A) \((\mathrm{x} / \alpha)+(\mathrm{y} / \beta)+(\mathrm{z} / \gamma)=1\) (B) \((\alpha / x)+(\beta / y)+(\gamma / z)=1\) (C) \(x+y+z=1\) (D) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\alpha \beta \gamma\)

For points \(\mathrm{A}(1,2,3), \mathrm{B}(5,4,1)\), the equation of plane which is perpendicular bisector of \(\underline{A B}\) is (A) \(x+2 y-7+z=0\) (B) \(2 x+y-z=7\) (C) \(x+2 y+z+7=0\) (D) \(2 x-2 y-z=7\)

The co-ordinates of the points of trisection of \(\underline{A B}\) is where \(\mathrm{A}(-5,7,2), \mathrm{B}(1,3,7)\) (A) \([-1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (B) \([1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (C) \([-1,4,(16 / 3)][-3,\\{(-11) / 2\\},\\{(-11) / 3\\}]\) (D) None of these

If the vertices of quadrilateral are \((1,1,1),(-2,4,1)\) \((-1,5,5),(2,2,5)\) then it is (A) rectangle (B) square (C) parallelogram (D) rhombus

Foot of perpendicular and perpendicular distance from \(\mathrm{P}(2,-1,5)\) and line \([(\mathrm{x}-1) / 10]=[(\mathrm{y}+2) /(-4)]=[(\mathrm{z}+8) /(-11)]\) is (A) \((-1,-2,3), \sqrt{14}\) (B) \((1,2,3), 14\) (C) \((-1,-2,-3), \sqrt{14}\) (D) \((1,2,3), \sqrt{14}\)
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