For the vectors a -3 -2 -4 k- and b -2 -5 - If a - x - y where x is parallel to b and y is perpendicular to bthen y- isA- 55 -22 -116 k-29B- 55 -22 -116 k-C- 55 -20 -116 k-29D- 44 -20 -116 k-29

The correct option is A 55î 22ĵ+116k̂29a=x+y ⇒a× nbsp;b=(x+y)× nbsp;b ⇒a× nbsp;b=y× nbsp;b ∵x is parallel to b ⇒ (a y)× nbsp;b=0 Let y=pî+qĵ+rk̂ Ta ...

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Byju's Answer

Standard XII

Physics

Subtraction of a Vector

For the vecto...

Question

For the vectors $\to a = 3^i + 2^j + 4^k$ and $\to b = 2^i + 5^j$ . If $\to a = \to x + \to y$ where $\to x$ is parallel to $\to b$ and $\to y$ is perpendicular to $\to b$ ,
then $\to y$ is

\frac{55^i - 22^j + 116^k}{29}

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55^i - 22^j + 116^k

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\frac{55^i - 20^j + 116^k}{29}

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\frac{44^i - 20^j + 116^k}{29}

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Solution

The correct option is A $\frac{55^i - 22^j + 116^k}{29}$
$\to a = \to x + \to y$
$\Rightarrow \to a \times \to b = (\to x + \to y) \times \to b$
$\Rightarrow \to a \times \to b = \to y \times \to b$
$∵ \to x is parallel to \to b$
$\Rightarrow (\to a - \to y) \times \to b = 0$

Let $\to y = p^i + q^j + r^k$
Taking their cross product
$∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 3 - p & 2 - q & 4 - r 2 & 5 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
Solving the above matrix we get
$r = 4 . . . (1)$
$5 p - 2 q = 11 . . . (2)$

Also $\to y$ is perpendicular to $\to b$ , so
$\to y . \to b = 0$
$(p^i + q^j + r^k) . (2^i + 5^j) = 0$
$\Rightarrow 2 p + 5 q = 0 . . . (3)$

From eqn $(2)$ and $(3)$
$q = \frac{- 22}{29}$ and $p = \frac{55}{29}$
$r = 4$

Hence $\to y = \frac{55^i - 22^j + 116^k}{29}$

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For the vectors →a=3^i+2^j+4^k and →b=2^i+5^j. If →a=→x+→y where →x is parallel to →b and →y is perpendicular to→b, then →y is

For the vectors $\to a = 3^i + 2^j + 4^k$ and $\to b = 2^i + 5^j$ . If $\to a = \to x + \to y$ where $\to x$ is parallel to $\to b$ and $\to y$ is perpendicular to $\to b$ ,
then $\to y$ is