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Let the vectors a→,b→,c→ given as a1i^+a2j^+a3k^,b1i^+b2j^+b3k^,c1i^+c2j^+c3k^ Then show that = a→×(b→+c→)=a→×b→+a→×c→. - Mathematics

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प्रश्न

Let the vectors `veca, vecb, vecc` given as `a_1hati + a_2hatj + a_3hatk, b_1hati + b_2hatj + b_3hatk, c_1hati + c_2hatj + c_3hatk` Then show that = `veca xx (vecb+ vecc) = veca xx vecb + veca xx vecc.`

योग

उत्तर

Given `a_1 hati + a_2hatj + a_3hatk, b_1hati + b_2hatj + b_3hatk, c_1hati + c_2hatj + c_3hatk`

To prove that `veca xx (vecb + vecc) = veca xx vecb + veca xx vecc`

`veca = a_1hati + a_2hatj + a_3hatk`

`vecb = b_1hati + b_2hatj + b_3hatk`

`vecc = c_1hati + c_2hatj + c_3hatk`

`(vecb + vecc) = (b_1 + c_1)hati + (b_2 + c_2)hatj + (b_3 + c_3)hatk`

`veca xx (vecb + vecc) = |(hati,  hatj,  hatk),(a_1,  a_2,  a_3), (b_1 + c_1, b_2 + c_2, b_3 + c_3)|`

`= hati [a_2b_3 + a_2c_3 - a_3b_2 - a_3c_2] + hatj [- a_1b_3 - a_1c_3 + a_3b_1 + a_3c_1] + hatk [a_1b_2 + a_1c_2 - a_2b_1 - a_2c_1]`       ....(i)

`veca xx vecb = |(hati, hatj, hatk), (a_1, a_2, a_3), (b_1, b_2, b_3)|`

`= hati [a_2b_3 - a_3b_2] + hatj [a_1b_3 - a_3b_1] + hatk [a_1b_2 - a_2b_1]`        ...(ii)

`veca xx vecc = |(hati, hatj, hatk), (a_1, a_2, a_3), (c_1, c_2, c_3)|`

`= hati [a_2c_3 - a_3c_2] + hatj [a_3c_1 - a_1 c] + hatk [a_1c_2 - a_2c_1]`        ...(iii)

Adding (ii) and (iii)

`hati [a_2c_3 - a_3c_2 + a_2b_3 - a_3b_2] + hatj [a_3c_1 - a_1c + a_1b_3 - a_3b_1] + hatk [a_1c_2 - a_2c_1 + a_1b_2 - a_2b_1]`   ....(iv)

From (i) and (iv)

`veca xx (vecb + vecc) = veca xx vecb + veca xx vecc`

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अध्याय 10: Vector Algebra - Exercise 10.4 [पृष्ठ ४५४]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 10 Vector Algebra
Exercise 10.4 | Q 7 | पृष्ठ ४५४

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