The scalar product of the vector a -k- with a unit vector along the sum of vectors b -2 -4 -5 k- and c -2 -3 k- is equalto one- Find the value of - and hence find the unit vector along b - c-

A=î+ĵ+k̂ b=2î+4ĵ 5k̂ c=λî+2ĵ+3k̂ b+c=(2+λ)î+6ĵ 2k̂ Unit vector along b+c=(2+λ)î+6ĵ 2k̂/√((2+λ)^2+6^2+( 2)^2)=v say Given,a.v=1 ⇒ (î+ĵ+k̂).[(2+λ )î+6ĵ 2k̂] ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

The scalar pr...

Question

The scalar product of the vector $\to a =^i +^j +^k$ with a unit vector along the sum of vectors $\to b = 2^i + 4^j - 5^k and \to c = λ^i + 2^j + 3^k$ is equal to one. Find the value of $λ$ and hence find the unit vector along $\to b + \to c$ .

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Solution

$\to a =^i +^j +^k$

$\to b = 2^i + 4^j - 5^k$

$\to c = λ^i + 2^j + 3^k$

$\to b + \to c = (2 + λ)^i + 6^j - 2^k$

Unit vector along $\to b + \to c = \frac{(2 + λ)^i + 6^j - 2^k}{\sqrt{(2 + λ)^{2} + 6^{2} + (- 2)^{2}}} = \to v say$

Given, $\to a . \to v = 1$

$\Rightarrow (^i +^j +^k) . \frac{[(2 + λ)^i + 6^j - 2^k]}{\sqrt{(2 + λ)^{2} + 6^{2} + 4}} = 1$

$\Rightarrow \frac{(2 + λ) .1 + 1. (6) + 1 (- 2)}{\sqrt{λ^{2} + 4 λ + 44}} = 1$

$\Rightarrow λ + 6 = \sqrt{λ^{2} + 4 λ + 44}$

$\Rightarrow λ^{2} + 12 λ + 36 = λ^{2} + 4 λ + 44$

$\Rightarrow 8 λ = 8$

$\Rightarrow λ = 1$

and unit vector along $\to b + \to c = \frac{3^i + 6^j - 2^k}{7}$
$= \frac{1}{7} (3^i + 6^j - 2^k) .$

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