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3rd Equation of Motion

A particle is...

Question

A particle is moving with initial velocity $\to u = ˆ i - ˆ j + ˆ k$ . What should be its acceleration so that it can remain moving in the same straight line?

A

\to a = 2 ˆ i - 2 ˆ j + 2 ˆ k

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B

\to a = - 2 ˆ i - 2 ˆ j + 2 ˆ k

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C

\to a = 3 ˆ i + 2 ˆ j + 3 ˆ k

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D

\to a = 1 ˆ i - 1 ˆ j

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Solution

The correct option is C $\to a = 2 ˆ i - 2 ˆ j + 2 ˆ k$
$u =^i -^j +^j$

$a = 2 (^i -^j +^k) = 2 u$

$a = 2^i - 2^j + 2^k$

The correct option is A.

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\to a = 2 i - 3 j, \to b = 2 j + 3 k,

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\to A = 3 i + 2 j - k

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Q. Given two vectors

\to a = - ˆ i + ˆ j + 2 ˆ k

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Q. Find the shortest distance between the following pairs of lines whose vector equations are:

(i)

\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k}) and \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})

\vec{r} = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 7 \hat{k}) and \vec{r} = - \hat{i} - \hat{j} - \hat{k} + μ (7 \hat{i} - 6 \hat{j} + \hat{k})

\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) and \vec{r} = (2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 5 \hat{k})

\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - t) \hat{k} and \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}

\vec{r} = (λ - 1) \hat{i} + (λ + 1) \hat{j} - (1 + λ) \hat{k} and \vec{r} = (1 - μ) \hat{i} + (2 μ - 1) \hat{j} + (μ + 2) \hat{k}

\vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 2 \hat{k}) and, \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})

\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} + \hat{j} - \hat{k} + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})

\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}

\vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})

[NCERT EXEMPLAR]

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