Show that the points A-B-C with position vectors a- -3- 4- 4k- - b- -2-k- and c- - 3- 5k- respectively-form the vertices of a right-angled triangle-

Given, a⃗=3î 4ĵ 4k̂ b⃗=2̂ ĵ+k̂ c⃗=î 3̂ĵ 5̂k̂ A⃗B⃗=b⃗ a⃗=(2̂î ĵ+k̂) (3̂î 4̂ĵ 4̂k̂) =2̂î ĵ+k̂ 3̂î+4̂ĵ+4̂k̂ =î+3̂ĵ+5̂k̂ B⃗C⃗=c⃗ ...

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Byju's Answer

Standard XII

Mathematics

Collinear Vectors

Show that the...

Question

Show that the points A,B,C with position vectors $\to a = (3^i - 4^j - 4^k)$ , $\to b = (2^i -^j +^k)$ and $\to c = (^i - 3^j - 5^k)$ respectively,form the vertices of a right-angled triangle.

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Solution

Given, $\to a = 3^i - 4^j - 4^k$
$\to b =^2 -^j +^k$
$\to c =^i -^3 j -^5 k$
$\to A B = \to b - \to a = (^2 i -^j +^k) - (^3 i -^4 j -^4 k)$
$=^2 i -^j +^k -^3 i +^4 j +^4 k$
$=^i +^3 j +^5 k$
$\to B C = \to c - \to b = (^i -^3 j -^5 k) - (^2 i -^j +^k)$
$=^i -^2 j -^6 k$
$\to C A = \to a - \to c = (^3 i -^4 j^4 k) - (^i -^3 j -^5 k)$
$=^3 i -^4 j -^4 k -^i +^3 j +^5 k$
$=^2 i -^j +^k$
$∴ \to A B . \to B C = (-^i +^3 j +^5 k) (-^i -^3 j -^5 k)$
=(-1)(-1)+(3)(-2)+(5)(-6)
$= 1 - 6 - 30 = - 35 \neq 0$
$\to A B . \to C A = (^- i +^- 3 j +^5 k) (^2 i^j +^k)$
=(-1)(2)+(3)(-1)+(5)(1)
=-2-3+5=0.
$∴ \to A B ⊥ \to C A$
$∴ Δ$ ABC froms a right angle triangle of A.

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Show that the points A,B,C with position vectors →a=(3^i−4^j−4^k), →b=(2^i−^j+^k) and →c=(^i−3^j−5^k) respectively,form the vertices of a right-angled triangle.

Show that the points A,B,C with position vectors $\to a = (3^i - 4^j - 4^k)$ , $\to b = (2^i -^j +^k)$ and $\to c = (^i - 3^j - 5^k)$ respectively,form the vertices of a right-angled triangle.