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Question 106 of 162

Q.(i) A force of magnitude 5 units acting parallel to 2i⃗−2j⃗+k⃗2\vec i - 2\vec j + \vec k displaces the point of application from (1,2,3)(1,2,3) to (5,3,7)(5,3,7). Find the work done by the force.

(ii) If A(−1,4,−3)A(-1,4,-3) is one end of a diameter AB of the sphere x2+y2+z2−3x−2y+2z−15=0x^2+y^2+z^2-3x-2y+2z-15=0, then find the coordinates of B.
Puducherry TnboardTamil Nadu HSC (DGE) Board 2016Subjective· 6mImportance★★★★★
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(i) Build the force vector from its magnitude and direction, then dot it with the displacement vector. (ii) Find the sphere's centre by completing the square, then use the midpoint formula for a diameter.

Part (i): Work done by the force

1. Unit vector along 2i⃗−2j⃗+k⃗2\vec i-2\vec j+\vec k.

∣2i⃗−2j⃗+k⃗∣=4+4+1=3 ⇒ u^=2i⃗−2j⃗+k⃗3|2\vec i-2\vec j+\vec k|=\sqrt{4+4+1}=3\ \Rightarrow\ \hat u=\frac{2\vec i-2\vec j+\vec k}{3}

2. Force vector of magnitude 5.

F⃗=5u^=103i⃗−103j⃗+53k⃗\vec F=5\hat u=\frac{10}{3}\vec i-\frac{10}{3}\vec j+\frac53\vec k

3. Displacement vector.

d⃗=(5−1)i⃗+(3−2)j⃗+(7−3)k⃗=4i⃗+j⃗+4k⃗\vec d=(5-1)\vec i+(3-2)\vec j+(7-3)\vec k=4\vec i+\vec j+4\vec k

4. Work done =F⃗⋅d⃗=\vec F\cdot\vec d.

W=103(4)+(−103)(1)+53(4)=403−103+203=503 unitsW=\frac{10}{3}(4)+\left(-\frac{10}{3}\right)(1)+\frac53(4)=\frac{40}{3}-\frac{10}{3}+\frac{20}{3}=\frac{50}{3}\text{ units}

Part (ii): The far end BB of the diameter

1. Find the centre of the sphere x2+y2+z2−3x−2y+2z−15=0x^2+y^2+z^2-3x-2y+2z-15=0 by completing the square: …

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