Question #26a9b

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Question #26a9b

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Answer 1 · 2017-06-26T15:21:46

drawn

Here positions of three points in 3D are given as (taking unit of length as $m$ $m$ )

$P \to (4, - 1, 5)$ $P->(4,-1,5)$

$P_{1} \to (- 1, - 1, 0)$ $P_1->(-1,-1,0)$

$P_{2} \to (3, - 1, - 3)$ $P_2->(3,-1,-3)$

So
$- - \to P P_{1} = - 5 ˆ i - 5 ˆ k$ $vec(PP_1)=-5hati-5hatk$

$- - \to P P_{2} = - ˆ i - 8 ˆ k$ $vec(PP_2)=-hati-8hatk$

$- -- \to P_{1} P_{2} = 4 ˆ i - 3 ˆ k$ $vec(P_1P_2)=4hati-3hatk$

If the positon of bird ,when it is on line $P_{1} P_{2}$ $P_1P_2$ be $O$ $O$ .

1st method

Let $O P_{1} : O P_{2} = m : n$ $OP_1:OP_2=m:n$

So $- - \to P O = \frac{n \cdot - - \to P P_{1} + m \cdot - - \to P P_{2}}{m + n}$ $vec(PO)=(n*vec(PP_1)+m*vec(PP_2))/(m+n)$

$\Rightarrow - - \to P O = \frac{n (- 5 ˆ i - 5 ˆ k) + m (- ˆ i - 8 ˆ k)}{m + n}$ $=>vec(PO)=(n(-5hati-5hatk)+m(-hati-8hatk))/(m+n)$

$\Rightarrow - - \to P O = \frac{(- 5 n - m) ˆ i + (- 5 n - 8 m) ˆ k}{m + n}$ $=>vec(PO)=((-5n-m)hati+(-5n-8m)hatk)/(m+n)$

Now $- - \to P O and - -- \to P_{1} P_{2}$ $vec(PO) and vec(P_1P_2)$ are perpendicularly inclined.

So $- - \to P O \cdot - -- \to P_{1} P_{2} = 0$ $vec(PO) *vec(P_1P_2)=0$

$\Rightarrow \frac{4 (- 5 n - m) - 3 (- 5 n - 8 m)}{m + n} = 0$ $=>(4(-5n-m)-3(-5n-8m))/(m+n)=0$

$\Rightarrow - 20 n - 4 m + 15 n + 24 m = 0$ $=>-20n-4m+15n+24m=0$

$\Rightarrow 20 m = 5 n$ $=>20m=5n$

$\Rightarrow \frac{m}{n} = \frac{1}{4}$ $=>m/n=1/4$

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2nd method

Then $- - \to P O$ $vec(PO)$ will be perpendicular to $- -- \to P_{1} P_{2}$ $vec(P_1P_2)$

So $- - \to O P_{1}$ $vec(OP_1)$ is the projection of $- - \to P P_{1}$ $vec(PP_1)$ on $- -- \to P_{2} P_{1}$ $vec (P_2P_1$

Hence $∣ ∣ ∣ - - \to O P_{1} ∣ ∣ ∣ = - -- \to P_{2} P_{1} \cdot - - \to P P_{1}$ $absvec(OP_1)=vec(P_2P_1)*vec(PP_1)$
$= \frac{(- 4 ˆ i + 3 ˆ k) \cdot (- 5 ˆ i - 5 ˆ k)}{∣ ∣ - 4 ˆ i + 3 k ∣ ∣}$ $=((-4hati+3hatk)*(-5hati-5hatk))/abs(-4hati+3k)$

$= \frac{20 - 15}{\sqrt{{(- 4)}^{2} + 3^{2}}} = 1$ $=(20-15)/sqrt((-4)^2+3^2)=1$

Again $- - \to O P_{2}$ $vec(OP_2)$ is the projection of $- - \to P P_{2}$ $vec(PP_2)$ on $- -- \to P_{1} P_{2}$ $vec (P_1P_2$

$∣ ∣ ∣ - - \to O P_{2} ∣ ∣ ∣ = - -- \to P_{1} P_{2} \cdot - - \to P P_{1}$ $absvec(OP_2)=vec(P_1P_2)*vec(PP_1)$
$= \frac{(4 ˆ i - 3 ˆ k) \cdot (- ˆ i - 8 ˆ k)}{∣ ∣ 4 ˆ i - 3 ˆ k ∣ ∣}$ $=((4hati-3hatk)*(-hati-8hatk))/abs(4hati-3hatk)$

$= \frac{- 4 + 24}{\sqrt{4^{2} + {(- 3)}^{2}}} = 4$ $=(-4+24)/sqrt(4^2+(-3)^2)=4$

So $- - \to P O$ $vec(PO)$ divides $P_{1} P_{2}$ $P_1P_2$ ar $O$ $O$ in the ratio $O P_{1} : O P_{2} = 1 : 4$ $OP_1:OP_2=1:4$

~~~~~~~~~~~~~~~~~~~~~~~~~~

Finally

$- - \to P O = \frac{4 \cdot - - \to P P_{1} + 1 \cdot - - \to P P_{2}}{1 + 4}$ $vec(PO)=(4*vec(PP_1)+1*vec(PP_2))/(1+4)$

$\Rightarrow - - \to P O = \frac{4 (- 5 ˆ i - 5 ˆ k) + 1 (- ˆ i - 8 ˆ k)}{5}$ $=>vec(PO)=(4(-5hati-5hatk)+1(-hati-8hatk))/5$

$\Rightarrow - - \to P O = \frac{- 20 ˆ i - 20 ˆ k - ˆ i - 8 ˆ k}{5}$ $=>vec(PO)=(-20hati-20hatk-hati-8hatk)/5$

$\Rightarrow - - \to P O = \frac{- 21 ˆ i - 28 ˆ k}{5}$ $=>vec(PO)=(-21hati-28hatk)/5$

$\Rightarrow ∣ ∣ ∣ - - \to P O ∣ ∣ ∣ = \frac{1}{5} ∣ ∣ - 21 ˆ i - 28 ˆ k ∣ ∣$ $=>abs(vec(PO))=1/5abs(-21hati-28hatk)$

$\Rightarrow ∣ ∣ ∣ - - \to P O ∣ ∣ ∣ = \frac{1}{5} \cdot \sqrt{{(- 21)}^{2} + {(- 28)}^{2}}$ $=>abs(vec(PO))=1/5*sqrt((-21)^2+(-28)^2)$

$\Rightarrow ∣ ∣ ∣ - - \to P O ∣ ∣ ∣ = \frac{1}{5} \cdot \sqrt{1225} = 7$ $=>abs(vec(PO))=1/5*sqrt(1225)=7$ m
Time (t) taken by the bird to cover the distance $P O$ $PO$ is

$t = \frac{distance PO}{velocity of the bird} = \frac{7 m}{2 m / s} = 3.5$ $t="distance PO"/"velocity of the bird"=(7m)/(2m"/"s)=3.5$ s

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