Math, asked by arnavpandit76, 7 hours ago

Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.​

Answers

Answered by mathdude500
3

\large\underline{\sf{Solution-}}

Given equation of lines are

\rm :\longmapsto\:x =  - y = kz

and

\rm :\longmapsto\:x - 2 = 2y + 1 =  - z + 1

Now, Consider,

\rm :\longmapsto\:x =  - y = kz

Divide whole equation by k, we get

\rm :\longmapsto\:\dfrac{x}{k}  = \dfrac{y}{ - k}  = \dfrac{z}{1}

can be further rewritten as

\rm :\longmapsto\:\dfrac{x - 0}{k}  = \dfrac{y - 0}{ - k}  = \dfrac{z - 0}{1}

\rm :\implies\:direction \: ratio \: of \: line \: is \: (k, - k,1)

In vector form direction ratios of line (1) is represented as

\bf\implies \:\vec{b_1} \:  =  \: k \hat{i} - k \hat{j} \:  +  \hat{k}

Now, Consider second equation of line,

\rm :\longmapsto\:x - 2 = 2y + 1 =  - z + 1

Divide whole equation by 2, we get

\rm :\longmapsto\:\dfrac{x - 2}{2}  = \dfrac{y + 0.5}{1}  = \dfrac{ - z + 1}{2}

can be rewritten as

\rm :\longmapsto\:\dfrac{x - 2}{2}  = \dfrac{y + 0.5}{1}  = \dfrac{ z - 1}{ - 2}

\rm :\implies\:direction \: ratio \: of \: line \: is \: (2, 1, - 2)

So,

In vector form, the direction ratios of the line is represented as

\bf\implies \:\vec{b_2} \:  =  \: 2 \hat{i} +  \hat{j} \: - 2\hat{k}

Now, further it is given that,

Line 1 and Line 2 are perpendicular to each other.

\rm :\implies\:\vec{b_1} \: . \: \vec{b_2} = 0

\rm :\longmapsto\:\: (k \hat{i} - k \hat{j} \:  +  \hat{k}) \: . \: (\: 2 \hat{i} +  \hat{j} \: - 2\hat{k}) = 0

\rm :\longmapsto\:2k - k  - 2 = 0

\rm :\longmapsto\:k  - 2 = 0

\bf\implies \:k = 2

Additional Information :-

Let us consider two lines in vector form as

\rm :\longmapsto\:\vec{r} = \vec{a_1} +  \alpha \vec{b_1}

and

\rm :\longmapsto\:\vec{r} = \vec{a_2} +   \beta  \vec{b_2}

then

1. Two lines are parallel iff

\boxed{ \rm \: \vec{b_1} = k\vec{b_2}}

2. Two lines are perpendicular iff

\boxed{ \rm \: \vec{b_1} \: . \: \vec{b_2} \:  =  \: 0}

3. Angle between two lines is

\boxed{ \rm \: cos \theta \:  =  \frac{\vec{b_1} \: . \: \vec{b_2}}{ |\vec{b_1}| \:  |\vec{b_2}|  } }

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