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Chapter 17: Problem 1602

\(\underline{\mathrm{a}}\) and \(\underline{\mathrm{b}}\) non coplanar. \(2 \underline{\mathrm{u}}-\underline{\mathrm{v}}=\underline{\mathrm{w}}\), if \(\underline{\mathrm{u}}=\mathrm{x} \underline{\mathrm{a}}+2 \mathrm{y} \underline{\mathrm{b}}\), \(\underline{v}=-2 y \underline{a}+3 x \underline{b}\) and \(\underline{w}=4 \underline{a}-2 \underline{b}\), find \(x\) and \(y=\) (a) \(\mathrm{x}=(8 / 7), \mathrm{y}=(2 / 7)\) (b) \(x=2, y=3\) (c) \(\mathrm{x}=(4 / 7), \mathrm{y}=(6 / 7)\) (d) \(x=(10 / 7), y=(4 / 7)\)

Short Answer

Expert verified
\( x = \frac{10}{7}, y = \frac{4}{7} \)

Step by step solution

01

Write vectors u, v, and w in terms of a and b

Using the given expressions, we have the following vectors: \( u = x\,a + 2y\,b \) \( v = -2y\,a + 3x\,b \) \( w = 4a - 2b \)
02

Express the vector equation in terms of a and b

We are given that: \( 2u - v = w \) Substitute the expressions for u, v, and w from Step 1: \( 2(x\,a + 2y\,b) - (-2y\,a + 3x\,b) = 4a - 2b \)
03

Simplify the equation

Expand and simplify the left side of the equation: \( 2x\,a + 4y\,b + 2y\,a - 3x\,b = 4a - 2b \) Combine the terms with a and b: \( (2x + 2y)\,a + (4y - 3x)\,b = 4a - 2b \)
04

Compare coefficients of a and b

Since the left and right side of the equation represents the same vector, we can say that the coefficients of a and b are equal: \( 2x + 2y = 4 \) \( 4y - 3x = -2 \)
05

Solve the system of linear equations

We have a system of two linear equations with two variables x and y: 1) \( 2x + 2y = 4 \) 2) \( 4y - 3x = -2 \) We can solve this system by first solving for y in equation 1: \( y = 2 - x \) Now, substitute this expression for y into equation 2: \( 4(2 - x) - 3x = -2 \)
06

Solve for x

Simplify and solve for x: \( 8 - 4x - 3x = -2 \) \( -7x = -10 \) \( x = \frac{10}{7} \)
07

Solve for y

Now, substitute the value of x back into the expression we found for y: \( y = 2 - \frac{10}{7} \) \( y = \frac{4}{7} \)
08

Check the answer in the given options

Comparing our solution with the given options, we find that: \( x = \frac{10}{7}, y = \frac{4}{7} \) This matches option (d).

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Most popular questions from this chapter

\(\underline{a}\) and \(\underline{b}\) are non coplanar and \(5 \underline{u}-3 \underline{\mathrm{v}}=\underline{\mathrm{w}} .\) if \(\underline{u}=\mathrm{ma}+2 \mathrm{nb}\) and \(\underline{\mathrm{v}}=-2 \mathrm{na}+3 \mathrm{mb}\) and \(\underline{\mathrm{w}}=4 \underline{\mathrm{a}}-2 \underline{\mathrm{b}} .\) find \(\mathrm{m}=\underline{\mathrm{n}}=\) (a) \((1 / 2),(1 / 4)\) (b) \((1 / 2),(1 / 3)\) (c) \((1 / 3),(1 / 4)\) (d) \(-(1 / 2),-(1 / 4)\)

if \(\underline{a} \perp \underline{b}\) then \(\underline{a} \times[\underline{a} \times\\{\underline{a} \times(\underline{a} \times \underline{b})\\}]=\) (a) \(-|\underline{a}|^{2} \underline{b}\) (b) \(|\underline{a}|^{4} \underline{b}\) (c) \(-|\underline{a}|^{6} \underline{b}\) (d) \(|a|^{6} \underline{b}\)

\(\underline{\mathrm{a}}\) and \(\underline{\mathrm{b}}\) are unit vector such as \(\underline{\mathrm{a}}+2 \underline{\mathrm{b}}\) and \(5 \underline{\mathrm{a}}-4 \underline{\mathrm{b}}\) are perpendicular, find the angle between \(\underline{a}\) and \(\underline{b}=\) (a) \(15^{\circ}\) (b) \(60^{\circ}\) (c) \(\cos ^{-1}(1 / 3)\) (d) \(\cos ^{-1}(2 / 7)\)

if \(\underline{\mathrm{OA}}=\underline{\mathrm{a}}, \underline{\mathrm{OB}}=\underline{\mathrm{b}}, \underline{\mathrm{OC}}=2 \underline{\mathrm{a}}+3 \underline{\mathrm{b}}, \underline{\mathrm{OD}}=\underline{\mathrm{a}}-2 \underline{\mathrm{b}}\). Length of \(\underline{\mathrm{OA}}\) is three times length of \(\underline{\mathrm{OB}} \& \mathrm{OA}\) is perpendicular to \(\underline{\mathrm{DB}}\) then \((\underline{\mathrm{BD}} \times \underline{\mathrm{AC}}) \cdot(\underline{\mathrm{OD}} \times \underline{\mathrm{OC}})\) is (a) \(7|\underline{\mathrm{a}} \times \underline{\mathrm{b}}|^{2}\) (b) 0 (c) \(7|\underline{a} \times \underline{b}|\) (d) \(42|\underline{\mathrm{a}} \times \underline{\mathrm{b}}|^{2}\)

\(\underline{\mathrm{a}}=(1,-1,0), \underline{\mathrm{b}}=(0,1,-1)\) and \(\underline{\mathrm{c}}=(-1,0,1)\) find the unit vector \(\underline{d}\) Such that \(\underline{\mathrm{a}} \cdot \underline{\mathrm{d}}=0=[\underline{\mathrm{b}} \underline{\mathrm{c}} \underline{\mathrm{d}}]\). (a) \(\pm(1 / \sqrt{6})(1,1,-2)\) (b) \(\pm(1 / \sqrt{3})(1,1,-1)\) (c) \(\pm(1 / \sqrt{3})(1,1,1)\) (d) \(\pm(0,0,1)\)
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