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Prove that the points `2hati-hatj+hatk, hati-3hatj-5hatk and 3hati-4hatj-4hatk ` are the vertices of a right angled triangle. Also find the remaining angles of the triangle. |
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Answer» We have, `vec(AB) =` (position of vector) - (position vectors of A) ` = (vecb - veca) = (2hati - hatj + hatk) - (3hati - 4hatj - 4hatk)` `= (-hati + 3hatj + 5hatk)`. `vec(BC) =` (position vector of C) - (position vector of B) `=(vecc - vecb) = (hati - 3hatj - 5hatk) - (2hati - hatk + hatk)`. `= (-hati - 2hatj - 6hatk)`. `vec(CA) =` (position of vector of A) - (position of vector B). ` = (veca - vecc) = (3hati - 4hatj - 4hatk) - (hati - 3hatj - 5hatk)`. `= (2hati - hatj + hatk)`. `:. |vec(AB)|^(2) - {(-1)^(2) + 3^(2) + 5^(2)} = (1+9+25) = 35`, `|vec(BC)|^(2) = {(-1)^(2) + (-2)^(2) + (-6)^(2) } = (1+4+36) = 41`. `|vec(CA)| = {2^(2) + (-1)^(2) + 1^(2)} = (4+1+1) = 6`. `:. |vec(AB)|^(2)= |vec(CA)|^(2) = |vec(BC)|^(2) rArr AB^(2) + CA^(2) = BC^(2)`. Hence, `DeltaABC` is right-angled at A. |
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