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Vector Addition

The position ...

Question

The position vectors of the points $A, B$ and $C$ are $i + j + k, 1 + 5 j - k$ and $2 i + 3 j + 5 k,$ respectively. The greatest angle of the triangle $A B C$ is

A

135^{\circ}

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B

90^{\circ}

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C

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

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D

{cos}^{- 1} \frac{5}{7}

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Solution

The correct option is B $90^{\circ}$
Given
$\to A =^i +^j +^k$
$\to B =^i + 5^j -^k$
$\to C = 2^i + 3^j + 5^k$
sides
$\to A B = \to B - \to A$
$\to A B =^i + 5^j -^k - (^i +^j +^k)$
$\to A B = 4^j - 2^k$
$∣ ∣ \to A B ∣ ∣ = \sqrt{4^{2} + 2^{2}} = \sqrt{20}$
$\to B C = \to C - \to B$
$\to B C = 2^i + 3^j + 5^k - (^i + 5^j -^k)$
$\to B C =^i - 2^j + 6^k$
$∣ ∣ \to B C ∣ ∣ = \sqrt{1^{2} + (- 2)^{2} + 6^{2}} = \sqrt{41}$
$\to C A = \to A - \to C$
$\to C A =^i +^j +^k - (2^i + 3^j + 5^k)$
$\to C A = -^i - 2^j - 4^k$
$∣ ∣ \to B C ∣ ∣ = \sqrt{(- 1)^{2} + (- 2)^{2} + (- 4)^{2}} = \sqrt{21}$
$\to A B \cdot \to B C = ∣ ∣ \to A B ∣ ∣ ∣ ∣ \to B C ∣ ∣ c o s α$
$(4^j - 2^k) \cdot (^i - 2^j + 6^k) = ∣ ∣ \to A ∣ ∣ ∣ ∣ \to B ∣ ∣ c o s α$
$- 8 - 12 = \sqrt{20} \sqrt{41} cos α$
$- 20 = \sqrt{20} \sqrt{41} cos α$
$cos α = \frac{- 20}{\sqrt{20} \sqrt{41}}$
$α = {cos}^{- 1} (- \sqrt{\frac{20}{41}})$
$\to B C \cdot \to C A = ∣ ∣ \to B C ∣ ∣ ∣ ∣ \to C A ∣ ∣ c o s β$
$(^i - 2^j + 6^k) \cdot (-^i - 2^j - 4^k) = \sqrt{41} \sqrt{21} c o s β$
$- 1 + 4 - 24 = \sqrt{21} \sqrt{41} cos β$
$cos β = - \frac{21}{\sqrt{21} \sqrt{41}}$
$β = {cos}^{- 1} (- \sqrt{\frac{21}{41}})$

$\to A B \cdot \to C A = ∣ ∣ \to A B ∣ ∣ ∣ ∣ \to C A ∣ ∣ cos β$
$(4^j - 2^k) \cdot (-^i - 2^j - 4^k) = \sqrt{20} \sqrt{21} cos γ$
$- 8 + 8 = \sqrt{20} \sqrt{41} cos γ$
$cos γ = 0$
$γ = 90^{.}$

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