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Chapter 16: Problem 1496

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

Short Answer

Expert verified
The value of \(a\) is 3, which corresponds to option (B).

Step by step solution

01

Identify the given vectors

Given vectors are \(\vec{A} = \mathrm{i} + \sqrt{3}\mathrm{j}\) and \(\vec{B} = \sqrt{3}\mathrm{i} + a\mathrm{j}\).
02

Calculate magnitudes of given vectors

Calculate the magnitude of \(\vec{A}\) and \(\vec{B}\). \(|\vec{A}| = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2\) \(|\vec{B}| = \sqrt{(\sqrt{3})^2 + a^2} = \sqrt{3 + a^2}\)
03

Use the dot product formula

Use the dot product formula with the angle between the two vectors, given as \(\pi/3\). \(\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos{\theta} = 2(\sqrt{3 + a^2})\cos{\frac{\pi}{3}}\) And calculate the dot product between \(\vec{A}\) and \(\vec{B}\) directly: \(\vec{A}\cdot\vec{B} = (1)(\sqrt{3}) + (\sqrt{3})(a) = \sqrt{3} + \sqrt{3}a\)
04

Set the two expressions for the dot product equal to each other

Set the dot product expressions equal to each other and solve for a: \(\sqrt{3} + \sqrt{3}a = 2(\sqrt{3 + a^2})\cos{\frac{\pi}{3}}\) Since \(\cos{\frac{\pi}{3}} = \frac{1}{2}\), the equation simplifies to: \(\sqrt{3} + \sqrt{3}a = (\sqrt{3 + a^2})\)
05

Solve for a

Square both sides to eliminate the square root: \(3 + 6a + 3a^2 = 3 + a^2\) Now, simplify the quadratic equation: \(2a^2 - 6a = 0\) Divide both sides by 2: \(a^2 - 3a = 0\) Factor out a: \(a(a - 3) = 0\) Solve for a: \(a = 0\) or \(a = 3\) Since we have two possible answers, let's check the options given to see which one matches. The correct answer is (B) 3.

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Most popular questions from this chapter

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) \((1,2,3), \sqrt{14}\)

If plane \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=-3\) express in form of \(x \cos \alpha+y \cos \beta+z \cos \gamma=p\), then perpendicular distance from origin to the plane is foot of perpendicular is and direction cosine is (A) \(1,[-(2 / 3),(2 / 3),-\overline{(1 / 3)}],-(2 / 3),(2 / 3),\\{(-1) / 3\\}\) (B) \(2,[-(2 / 3),(2 / 3),-(1 / 3)],-(2 / 3),(2 / 3),\\{(-1) / 3\\}\) (C) \(1,[(2 / 3),(2 / 3),-(1 / 3)],(2 / 3),(2 / 3),(1 / 3)\) (D) None of these

If angle between two unit vectors \(\underline{a} \& \underline{b}\) is \(\alpha\), then \(|\underline{\mathrm{a}}-\underline{\mathrm{b}} \cos \alpha|=0<\alpha<(\pi / 2)\) (A) \(\sin \alpha\) (B) \(\sin (\alpha / 2)\) (C) \(\sin 2 \alpha\) (D) \(\sin ^{2}(\alpha / 2)\)

\([(2-3 \mathrm{x}) / 6]=[(\mathrm{y}+1) / 2]=[(1-z) /(-2)]\) direction of line \(\overline{(A)}+2,2,2\) (B) \(-1,1,1\) (C) \(-3,2,2\) (D) \(6,2,-2\)

Line \(L: \underline{r}=(8,-9,10)+k(3,-16,7), \mathrm{k} \in R\) and \(\mathrm{M}: \underline{\mathrm{r}}=(15,29,5)+\mathrm{k}(3,8,-5), \mathrm{k} \in \mathrm{R} .\) If \(\mathrm{P} \in \mathrm{L}, \mathrm{Q} \in \mathrm{M}\), where \(\underline{P Q}\) is shortest distance between \(L\) and \(M\) then \(P Q=\) (A) \(\sqrt{14}\) (B) 14 (C) \((1 / 14)\) (D) \((1 / \sqrt{14})\)
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