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Maths-

General

Easy

Question

A vector $n$ of magnitude 8 units is inclined to x-axis at $4 5 to the power of o end exponent$ , y-axis at $6 0 to the power of o end exponent$ and an acute angle with z-axis. If a plane passes through a point $left parenthesis square root of 2 comma negative 11 right parenthesis$ and is normal to $n$ , then its equation in vector form is

$r . (\sqrt{2} i + j + k) = 4$
$r . (\sqrt{2} i + j + k) = 2$
$r . (i + j + k) = 4$
None of these

The correct answer is: $r. left parenthesis square root of 2 i plus j plus k right parenthesis equals 2$

Let $gamma$ be the angle made by $n$ with z-axis.
Then direction cosines of $n$ are $l equals cos invisible function application 4 5 to the power of o end exponent equals fraction numerator 1 over denominator square root of 2 end fraction comma$ $m equals cos invisible function application 6 0 to the power of o end exponent equals fraction numerator 1 over denominator 2 end fraction$ and $n equals cos invisible function application gamma$ .
\ $l to the power of 2 end exponent plus m to the power of 2 end exponent plus n to the power of 2 end exponent equals 1 rightwards double arrow open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator 1 over denominator 2 end fraction close parentheses to the power of 2 end exponent plus n to the power of 2 end exponent equals 1$
Þ $n to the power of 2 end exponent equals fraction numerator 1 over denominator 4 end fraction rightwards double arrow n equals fraction numerator 1 over denominator 2 end fraction$ , [ $because gamma$ is acute, \ $n equals cos invisible function application gamma greater than 0$ ]
We have $vertical line n vertical line equals 8$ , \ $n equals vertical line n vertical line left parenthesis l i plus m j plus n k right parenthesis$
$rightwards double arrow n equals 8 open parentheses fraction numerator 1 over denominator square root of 2 end fraction i plus fraction numerator 1 over denominator 2 end fraction j plus fraction numerator 1 over denominator 2 end fraction k close parentheses equals 4 square root of 2 i plus 4 j plus 4 k$
The required plane passes through the point $left parenthesis square root of 2 comma negative 11 right parenthesis$ having position vector $a equals square root of 2 i minus j plus k$ .
So, its vector equation is $left parenthesis r minus a right parenthesis. n equals 0$ or $r. n equals a. n$
Þ $r. left parenthesis 4 square root of 2 i plus 4 j plus 4 k right parenthesis equals left parenthesis square root of 2 i minus j plus k right parenthesis. left parenthesis 4 square root of 2 i plus 4 j plus 4 k right parenthesis$
Þ $r. left parenthesis square root of 2 i plus j plus k right parenthesis equals 2$ .

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Chemistry-General

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parallel

The arrangement of the following in ascending order of angle of rotation to eliminate xy term in the following equations.
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parallel

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parallel

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parallel

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