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Chapter 2: Problem 292

If resultant of $\mathrm{A}^{\rightarrow}=2 \hat{1}+\hat{\jmath}-\mathrm{k}^{\wedge}, \mathrm{B}^{\rightarrow}=\hat{\imath}-2 \hat{\mathrm{j}}+3 \mathrm{k}^{\text {and }} \mathrm{C}^{\rightarrow}$ is unit vector in y direction, then \(\mathrm{C}^{\rightarrow}\) is (A) \(-\hat{j}\) (B) \(3 \hat{i}-2 \hat{j}+2 \mathrm{k}^{\wedge}\) (C) (D) \(2 \hat{i}+3 \mathrm{k}^{\wedge}\)

Short Answer

Expert verified
Therefore, the short answer is: \(\mathrm{C}^{\rightarrow} = -\hat{j}\).

Step by step solution

01

Find the resultant vector#\tag_content#Add vectors \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) to find the resultant vector \(\mathrm{R}^{\rightarrow}\): \[\mathrm{R}^{\rightarrow} = \mathrm{A}^{\rightarrow} + \mathrm{B}^{\rightarrow}\]

Step 2: Calculate \(\mathrm{R}^{\rightarrow}\)#\tag_content#Now, substitute the given values of \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) into the equation and calculate the resultant vector: \[\mathrm{R}^{\rightarrow} = (2\hat{\imath} + \hat{\jmath} - \hat{k}) + (\hat{\imath} - 2\hat{\jmath} + 3\hat{k})\]
02

Simplify the resultant vector#\tag_content#Combine the like terms (i.e. components in the same direction) to simplify the resultant vector: \[\mathrm{R}^{\rightarrow} = (2\hat{\imath} + \hat{\imath}) + (\hat{\jmath} - 2\hat{\jmath}) + (-\hat{k} + 3\hat{k})\] \[\mathrm{R}^{\rightarrow} = 3\hat{\imath} - \hat{\jmath} + 2\hat{k}\]

Step 4: Identify \(\mathrm{C}^{\rightarrow}\) as a unit vector in the y direction#\tag_content#According to the given information, \(\mathrm{C}^{\rightarrow}\) is a unit vector in the y direction and also the resultant of the two given vectors, meaning it is equal to the previously calculated \(\mathrm{R}^{\rightarrow}\). The only option among (A), (B), (C), and (D) which is a unit vector in the y direction, is: (A) \(-\hat{j}\) Thus, the correct option is (A), and \(\mathrm{C}^{\rightarrow} = -\hat{j}\).

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Most popular questions from this chapter

\(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\) and \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{~A}^{-}+\mathrm{B}^{-}\right|\) What is the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) \((\mathrm{A})(\pi / 6)\) (B) \((5 \pi / 6)\) (C) \((2 \pi / 3)\) (D) \((\pi / 3)\)

Stone \(\mathrm{A}\) is thrown in horizontal direction with velocity of $10 \mathrm{~ms}^{-1}\( at the same time stone \)\mathrm{B}$ freely falls vertically in downward direction. Calculate the velocity of \(\mathrm{B}\) with respect to \(\mathrm{A}\) after 10 second. (A) \(10 \mathrm{~ms}^{-1}\) (B) \(\sqrt{(101)} \mathrm{ms}^{-1}\) (C) \(10 \sqrt{(101)} \mathrm{ms}^{-1}\) (D) 0

A freely falling stone crashes through a horizontal glass plate at time \(t\) and losses half of its velocity. After time \((t / 2)\) it falls on the ground. The glass plate is \(60 \mathrm{~m}\) high from the ground. Find the total distance travelled by the stone. \(\left[g=10 \mathrm{~ms}^{-2}\right]\) (A) \(120 \mathrm{~m}\) (B) \(80 \mathrm{~m}\) (C) \(100 \mathrm{~m}\) (D) \(140 \mathrm{~m}\)

Comprehensions type questions. A particle is moving in a circle of radius \(R\) with constant speed. The time period of the particle is \(\mathrm{T}\) Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Range of a projectile is \(R\) and maximum height is \(\mathrm{H}\). Find the area covered by the path of the projectile and horizontal line. (A) \((2 / 3) \mathrm{RH}\) (B) \((5 / 3) \mathrm{RH}\) (C) \((3 / 5) \mathrm{RH}\) (D) \((6 / 5) \mathrm{RH}\)

$\quad\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)=\ldots$ (A) 0 (B) \(A^{2}+B^{2}\) (C) \(\sqrt{\left(A^{2}+B^{2}\right)}\) (D) \(\mathrm{A}^{2} \mathrm{~B}^{2}\)
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