Let c-x-y-zk- is the external angle bisector between two vectors a-2-3k- and b-2-3k- -c-3-46- Then x-y-z-

Given vectors, nbsp;a=2î ĵ+3k̂ and b=î+2ĵ 3k̂, nbsp; external angle bisector is λ(â b̂) c=λ(2î ĵ+3k̂√(14) î+2ĵ 3k̂√(14)) =λ√(14)(î 3ĵ+6k̂) ⇒ |c|=√(46)λ√(14 ...

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Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

Let c=xî+yĵ+z...

Question

Let $\to c = x^i + y^j + z^k$ is the external angle bisector between two vectors $\to a = 2^i -^j + 3^k$ and $\to b =^i + 2^j - 3^k, | \to c | = 3 \sqrt{46} .$ Then $x + y + z =$

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Solution

Given vectors, $\to a = 2^i -^j + 3^k$ and $\to b =^i + 2^j - 3^k,$
external angle bisector is $λ (^a -^b)$
$\to c = λ (\frac{2^i -^j + 3^k}{\sqrt{14}} - \frac{^i + 2^j - 3^k}{\sqrt{14}}) = \frac{λ}{\sqrt{14}} (^i - 3^j + 6^k) \Rightarrow | \to c | = \frac{\sqrt{46} λ}{\sqrt{14}} = 3 \sqrt{46} \Rightarrow λ = 3 \sqrt{14} \Rightarrow \to c = 3^i - 9^j + 18^k ∴ x + y + z = 3 - 9 + 18 = 12$

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Condition for Two Lines to be on the Same Plane

Standard XII Mathematics

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Let →c=x^i+y^j+z^k is the external angle bisector between two vectors →a=2^i−^j+3^k and →b=^i+2^j−3^k,|→c|=3√46. Then x+y+z=

Given vectors, →a=2^i−^j+3^k and →b=^i+2^j−3^k, external angle bisector is λ(^a−^b) →c=λ(2^i−^j+3^k√14−^i+2^j−3^k√14)=λ√14(^i−3^j+6^k)⇒|→c|=√46λ√14=3√46⇒λ=3√14⇒→c=3^i−9^j+18^k∴x+y+z=3−9+18=12

Let $\to c = x^i + y^j + z^k$ is the external angle bisector between two vectors $\to a = 2^i -^j + 3^k$ and $\to b =^i + 2^j - 3^k, | \to c | = 3 \sqrt{46} .$ Then $x + y + z =$