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Question 60 of 68

Q.A bird flies through a distance in a straight line given by the vector ๐‘–ฬ‚ + 2๐‘—ฬ‚ + ๐‘˜ฬ‚ . A man standing beside a straight metro rail track given by ๐‘Ÿโƒ— = (3 + ฮป)๐‘–ฬ‚ + (2ฮป โˆ’ 1)๐‘—ฬ‚ + 3ฮป๐‘˜ฬ‚ is observing the bird. The projected length of its flight on the metro track is
(A) 6 โˆš14 units
(B) 14 โˆš6 units
(C) 8 โˆš14 units
(D) 5 โˆš6 units

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The projected length is the scalar projection of the flight vector onto the track direction: 814=4147โ‰ˆ2.14\dfrac{8}{\sqrt{14}} = \dfrac{4\sqrt{14}}{7} \approx 2.14 units, which does not equal any of the four printed options.

The idea

The bird's flight is a vector; the metro track is a straight line with a fixed direction. The "projected length of the flight on the track" is the length of the shadow the flight vector casts along the track, i.e. the scalar projection of the flight vector onto the track's direction vector.

Set up

Flight vector:

aโƒ—=i^+2j^+k^.\vec{a} = \hat{i} + 2\hat{j} + \hat{k}.

The track is rโƒ—=(3+ฮป)i^+(2ฮปโˆ’1)j^+3ฮปk^\vec{r} = (3+\lambda)\hat{i} + (2\lambda-1)\hat{j} + 3\lambda\hat{k}. Splitting the fixed part from the ฮป\lambda part,

rโƒ—=(3i^โˆ’j^)+ฮป(i^+2j^+3k^),\vec{r} = (3\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k}),

so the track's direction is

bโƒ—=i^+2j^+3k^.\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}.

Only the direction matters for a projection; the constant part just fixes where the line sits.

Work the steps

1. Dot product.

aโƒ—โ‹…bโƒ—=(1)(1)+(2)(2)+(1)(3)=8.\vec{a}\cdot\vec{b} = (1)(1) + (2)(2) + (1)(3) = 8.

2. Length of the direction.

โˆฃbโƒ—โˆฃ=12+22+32=14.|\vec{b}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}.

3. Scalar projection. โ€ฆ

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