If a -2 -2 k- and b -7 -2 -3 k- then express b in the form of b - b 1- b 2- where b 1 is parallel to a and b 2 isperpendicular to a-

Since b_1 is parallel to a ⇒b_1 = λa ⇒b_1 = 2 λî λĵ 2 λk̂ Also, if nbsp; b = b_1 + b_2 ⇒ 7 î + 2 ĵ 3 k̂ = 2 λî λĵ 2 λk̂ + b_2 ⇒b_2 = (7 2 λ ) î + ( ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If a =2 î-ĵ-2...

Question

If $\to a = 2^i -^j - 2^k$ and $\to b = 7^i + 2^j - 3^k,$ then express $\to b$ in the form of $\to b = {\to b}_{1} + {\to b}_{2},$ where ${\to b}_{1}$ is parallel to $\to a$ and ${\to b}_{2}$ is perpendicular to $\to a .$

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Solution

Since ${\to b}_{1}$ is parallel to $\to a$
$\Rightarrow {\to b}_{1} = λ \to a \Rightarrow {\to b}_{1} = 2 λ^i - λ^j - 2 λ^k$

Also, if
$\to b = {\to b}_{1} + {\to b}_{2} \Rightarrow 7^i + 2^j - 3^k = 2 λ^i - λ^j - 2 λ^k + {\to b}_{2} \Rightarrow {\to b}_{2} = (7 - 2 λ)^i + (2 + λ)^j + (2 λ - 3)^k$

It is given that
${\to b}_{2} ⊥ \to a \Rightarrow (7 - 2 λ) (2) + (2 + λ) (- 1) + (2 λ - 3) (- 2) = 0 \Rightarrow 14 - 4 λ - 2 - λ - 4 λ + 6 = 0 \Rightarrow - 9 λ + 18 = 0 \Rightarrow λ = - 2 ∴ {\to b}_{1} = - 4^i + 2^j + 4^k$
${\to b}_{2} = 11^i + 0^j - 7^k ∴ 7^i + 2^j - 3^k = (- 4^i + 2^j + 4^k) + (11^i + 0^j - 7^k)$

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If →a=2^i−^j−2^k and →b=7^i+2^j−3^k, then express →b in the form of →b=→b1+→b2, where →b1 is parallel to →a and →b2 is perpendicular to →a.

If $\to a = 2^i -^j - 2^k$ and $\to b = 7^i + 2^j - 3^k,$ then express $\to b$ in the form of $\to b = {\to b}_{1} + {\to b}_{2},$ where ${\to b}_{1}$ is parallel to $\to a$ and ${\to b}_{2}$ is perpendicular to $\to a .$