Q10 -160-AB-3i-j-k-AC-i-j-3k-if -160-points -160-P -160-on -160-the -160-line -160-segment -160-BC -160-is -160-equidistant -160-from -160-AB -8201-and -160-Ac -160-tha -160-AP -160-is

Dear student A #160;point #160;equidistant #160;from #160;AB #8594; #160;and #160;AC #8594; #160;is #160;the #160;bisector #160;of #16 ...

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Standard IX

Mathematics

Distance Formula

Q10 #160;AB...

Question

$Q 10 A B = 3 \hat{i} + \hat{j} + \hat{k} A C = \hat{i} - \hat{j} + 3 \hat{k} i f p o i n t s P o n t h e l i n e s e g m e n t B C i s e q u i d i s \tan t f r o m A B a n d A c t h a A P i s$

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Solution

Dear student
$A point equidistant from \vec{AB} and \vec{AC} is the bisector of ∠ BAC . A vector along the internal ∠ bisector of ∠ BAC = \frac{\vec{AB}}{|\vec{AB}|} + \frac{\vec{AC}}{|\vec{AC}|} = \frac{3 i + j - k}{\sqrt{3^{2} + 1^{2} + {(- 1)}^{2}}} + \frac{i - j + 3 k}{\sqrt{1^{2} + {(- 1)}^{2} + 3^{2}}} = \frac{3 i + j - k}{\sqrt{11}} + \frac{i - j + 3 k}{\sqrt{11}} = \frac{1}{\sqrt{11}} (4 i + 2 k) So, \vec{AP} = t (2 i + k) and \vec{BP} = \vec{AP} - \vec{AB} = t (2 i + k) - (3 i + j - k) = (2 t - 3) i - j + (t + 1) k Also \vec{BC} = \vec{AC} - \vec{AB} = (i - j + 3 k) - (3 i + j - k) = - 2 i - 2 j + 4 k But \vec{BP} = s \vec{BC} So, (2 t - 3) i - j + (t + 1) k = s (- 2 i - 2 j + 4 k) So, (2 t - 3) = - 2 s, - 1 = - 2 s, t + 1 = 4 s So, s = \frac{1}{2} and t = 1 So, \vec{AP} = 2 i + k$
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Similar questions

A B = 3 i + j - k

and

A C = i - j + 3 k

. If the point

P

on the line segment

B C

is equidistant from

A B

and

A C,

then

A P

Q. Let

- - \to A B = 3 i + j - k

and

- - \to A C = i - j + 3 k

. If the point

P

on the line segment

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is equidistant from

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and

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\to p

Q. Let

- - \to A B = 3^i +^j -^k

and

- - \to A C =^i -^j + 3^k

and a point

P

on the line segment

B C

be equidistant from

A B

and

A C

, the

- - \to A P

A B = 3^i +^j -^k

and

A C =^i -^j + 3^k .

If the point

P

on the line segment

B C

is equdistant from

A B

and

A C,

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Q. Using ruler and compasses only, construct an isosceles

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and

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(ii)

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Q10 AB=3i^+j^+k^AC=i^-j^+3k^if points P on the line segment BC is equidistant from AB and Ac tha AP is

$Q 10 A B = 3 \hat{i} + \hat{j} + \hat{k} A C = \hat{i} - \hat{j} + 3 \hat{k} i f p o i n t s P o n t h e l i n e s e g m e n t B C i s e q u i d i s \tan t f r o m A B a n d A c t h a A P i s$