The position vectors of the point A- B and C are 2-k- -3-5k- and 3-4-4k- The greatest angle of triangle ABC is-

The correct option is A 90^0 A⃗B⃗= î 2ĵ 6k̂ ...(i) B⃗C⃗=2î ĵ+k̂ ...(ii) C⃗A⃗= î+3ĵ+5k̂ ...(iii)Now B⃗C⃗.C⃗A⃗=(2î ĵ+k̂).( î+3ĵ+5k̂) ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

The position ...

Question

The position vectors of the point $A, B$ and $C$ are $2^i -^j +^k,^i - 3^j - 5^k$ and $3^i - 4^j - 4^k .$ The greatest angle of triangle $A B C$ is.

90^{0}

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120^{0}

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135^{0}

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{cos}^{- 1} (\frac{2}{3})

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Solution

The correct option is A $90^{0}$
$\to A B = -^i - 2^j - 6^k$ ...(i)
$\to B C = 2^i -^j +^k$ ...(ii)
$\to C A = -^i + 3^j + 5^k$ ...(iii)
Now $\to B C . \to C A = (2^i -^j +^k) . (-^i + 3^j + 5^k)$
$= - 2 - 3 + 5$
$= 0$
Hence $B C$ is perpendicular to $C A$ . Therefore $A B C$ is a right angled triangle.
Hence $∠ C$ is $90^{0}$ .
Thus the greatest angle is $90^{0}$

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The position vectors of the point A,B and C are 2^i−^j+^k, ^i−3^j−5^k and 3^i−4^j−4^k. The greatest angle of triangle ABC is.

The correct option is A 900→AB=−^i−2^j−6^k ...(i)→BC=2^i−^j+^k...(ii)→CA=−^i+3^j+5^k ...(iii)Now →BC.→CA=(2^i−^j+^k).(−^i+3^j+5^k)=−2−3+5=0Hence BC is perpendicular to CA. Therefore ABC is a right angled triangle. Hence ∠C is 900.Thus the greatest angle is 900

The position vectors of the point $A, B$ and $C$ are $2^i -^j +^k,^i - 3^j - 5^k$ and $3^i - 4^j - 4^k .$ The greatest angle of triangle $A B C$ is.