Question

position vectors of A and B with respect to origin O be 2î-j+k and 5i+3j+k respectively then find ABvec. and|AB|​

Answer 1

Answer:

Solution

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We have,

A=−2i+j−k

B=−4i+2j+2k

C=6i−3j−13k

Now,

AB=Position vector of B−Position vector of A

=(−4i+2j+2k)−(−2i+j−k)

=−2i+j+3k

Similarly,

AC=Position vector of C−Position vector of A

=(6i−3j−13k)−(−2i+j−k)

=8i−4j−12k

Since,

AB=λAC

−2i+j+3k=λ(8i−4j−12k)

λ=−

4

1

Step-by-step explanation:

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Answer 2

Answer:

|vec(OA)|= |vec(OB)| = sqrt(14)

<br>

Delta AOB

is isosceles. Hence, the bisector of angle

AOB

will bisect the base

AB

. <br> Hence, P is the midpoint

(2, 2, -2)

of AB. Therefore, <br>

" "vec(OP)= 2(hati+hatj-hatk)