Let c be a unit vector which is coplanar with a -2 k- and b -2 -k- such that c is perpendicular to a- If P be the lengthprojection of c along b- then the value of 11- P 2 is

Let c = x a + y b ⇒ nbsp; c = î (x+2y) + ĵ ( x y) +k̂(2x + y ) Now, c . a = 0 ⇒ x+2y +x+y+4x+2y = 0 ⇒ y = 6x5 c = 7x5î + x5ĵ nbsp; +4x5k̂ We know that, | ...

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Byju's Answer

Standard XII

Physics

Acceleration in 2D

Let c be a un...

Question

Let $\to c$ be a unit vector which is coplanar with $\to a =^i -^j + 2^k$ and $\to b = 2^i -^j +^k$ such that $\to c$ is perpendicular to $\to a$ . If $P$ be the length projection of $\to c$ along $\to b$ , then the value of $\frac{11}{P^{2}}$ is

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Solution

Let $\to c = x \to a + y \to b$
$\Rightarrow \to c =^i (x + 2 y) +^j (- x - y) +^k (2 x + y)$
Now,
$\to c . \to a = 0 \Rightarrow x + 2 y + x + y + 4 x + 2 y = 0 \Rightarrow y = - \frac{6 x}{5}$
$\to c = \frac{- 7 x}{5}^i + \frac{x}{5}^j + \frac{4 x}{5}^k$
We know that,
$| \to c | = 1 \Rightarrow \frac{49 x^{2} + x^{2} + 16 x^{2}}{25} = 1 \Rightarrow x^{2} = \frac{25}{66} \Rightarrow \to c = \pm \frac{5}{\sqrt{66}} (\frac{- 7}{5}^i + \frac{1}{5}^j + \frac{4}{5}^k)$
The perpendicular distance will be,
$P = \frac{\to c \cdot \to b}{| \to b |} = \frac{\sqrt{11}}{6} \Rightarrow \frac{11}{P^{2}} = 36$

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Let →c be a unit vector which is coplanar with →a=^i−^j+2^k and →b=2^i−^j+^k such that →c is perpendicular to →a. If P be the length projection of →c along →b, then the value of 11P2 is

Let →c=x→a+y→b ⇒→c=^i(x+2y)+^j(−x−y)+^k(2x+y) Now, →c.→a=0⇒x+2y+x+y+4x+2y=0⇒y=−6x5 →c=−7x5^i+x5^j+4x5^k We know that, |→c|=1⇒49x2+x2+16x225=1⇒x2=2566⇒→c=±5√66(−75^i+15^j+45^k) The perpendicular distance will be, P=→c⋅→b|→b|=√116⇒11P2=36

Let $\to c$ be a unit vector which is coplanar with $\to a =^i -^j + 2^k$ and $\to b = 2^i -^j +^k$ such that $\to c$ is perpendicular to $\to a$ . If $P$ be the length projection of $\to c$ along $\to b$ , then the value of $\frac{11}{P^{2}}$ is