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-

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Angle between Pair of Straight Lines

If a = 2 i ...

Question

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

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Solution

Given that $a = 2 i - j - 2 k$ and $b = 7 i + 2 j - 3 k$
Given that $b = b_{1} + b_{2}$ where $b_{1}$ is parallel to $a$ and $b_{2}$ is perpendicular to $a$
Let $b_{1} = k a$ , we have $a . b = a . b_{1} + a . b_{2} = k a . a + 0 = k a . a$
$\Rightarrow k = \frac{a . b}{a . a} = \frac{14 - 2 + 6}{4 + 1 + 4} = 2$
So we get $b_{1} = 4 i - 2 j - 4 k$
Now we have $b \times a = b_{1} \times a + b_{2} \times a = b_{2} \times a$
Let $b_{2} = x i + y j + z k$ , we get $- 7 i + 8 j - 11 k = (- 2 y + z) i + j (2 x + 2 z) + k (- x - 2 y)$
By comparing we get $2 y - z = 7$ , $x + z = 4$ and $x + 2 y = 11$
By solving , we get $x = 3, y = 4, z = 1$
So we get $b_{2} = 3 i + 4 j + k$

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If →a=2ˆi−ˆj−2ˆk and →b=7ˆi+2ˆj−3ˆk, then express →b in the form of →b=→b1+→b2, where →b1 is parallel to →a and →b2 is perpendicular to →a.

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