The plane {r}=s({i}+2{j}-4k)+t(3i+4{j}-4k)+(1-t)(2{i}-7{j}-3{k}) is parallel to the line

Correct Answer is : r⃗=(−i^+j^−k^)+t(−i^−2j^+4k^){r}=(-{i}+{j}-{k})+t(-{i}-2{j}+4{k})r=(−i^+j^−k^)+t(−i^−2j^+4k^)

{r}=(-{i}+{j}-{k})+t(-{i}-2{j}+4{k})

{r}=(-{i}+{j}-{k})+t({i}-2{j}+4{k})

{r}=({i}+{j}-{k})+t(-{i}-4{j}+7{k})

{r}=(-{i}+{j}-{k})+t(-2{i}+2{j}+4{k})

{r}=(-{i}+{j}-3k)+t(2{i}+6{j}-8{k})

Correct Answer is : r⃗=(−i^+j^−k^)+t(−i^−2j^+4k^){r}=(-{i}+{j}-{k})+t(-{i}-2{j}+4{k})r=(−i^+j^−k^)+t(−i^−2j^+4k^)

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KEAM 2010 Math Paper

Question : 15 of 120

Marks: +1, -0

The plane

\vec{r}=s(\hat{i}+2\hat{j}-4k)+t(3i+4\hat{j}-4k)+(1-t)(2\hat{i}-7\hat{j}-3\hat{k})

is parallel to the line

$\vec{r}=(-\hat{i}+\hat{j}-\hat{k})+t(-\hat{i}-2\hat{j}+4\hat{k})$
$\vec{r}=(-\hat{i}+\hat{j}-\hat{k})+t(\hat{i}-2\hat{j}+4\hat{k})$
$\vec{r}=(\hat{i}+\hat{j}-\hat{k})+t(-\hat{i}-4\hat{j}+7\hat{k})$
$\vec{r}=(-\hat{i}+\hat{j}-\hat{k})+t(-2\hat{i}+2\hat{j}+4\hat{k})$
$\vec{r}=(-\hat{i}+\hat{j}-3k)+t(2\hat{i}+6\hat{j}-8\hat{k})$

Solution:

Given plane is

\Rightarrow \vec{r}=r(\hat{i}+2\hat{j}-4\hat{k})+t(3\hat{i}+4\hat{j}-4\hat{k})

+(1-t)(2\hat{i}-7\hat{j}-3\hat{k})

\Rightarrow \vec{r}=(2\hat{i}-7\hat{j}+3\hat{k})+s(\hat{i}+2\hat{j}-4\hat{k})

+t(\hat{i}+11\hat{j}-\hat{k})

Comparing it with the equation of plane

\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c},

we get

\vec{b}=\hat{i}+2\hat{j}-4\hat{k}

and

\vec{c}=\hat{i}+11\hat{j}-\hat{k}

Now,

\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -4 \\ 1 & 11 & -1 \end{vmatrix}

=42\hat{i}-3\hat{j}+9\hat{k}

\therefore

Parametric form of plane is

\vec{r}\cdot(\vec{b}\times \vec{c})=\vec{a}\cdot(\vec{b}\times \vec{c})

\Rightarrow \vec{r}\cdot (42\hat{i}-3\hat{j}+9\hat{k})

=(2\hat{i}-7\hat{j}+3\hat{k})\cdot(42\hat{i}-3\hat{j}+9\hat{k})

which is of the form

\vec{r}\cdot \vec{r}=d

\Rightarrow\;\vec{r}=42\hat{i}-3\hat{j}+9\hat{k}

Now, the line given in option (a) is

\vec{r}=(-\hat{i}+\hat{j}+\hat{k})+t(-\hat{i}-2\hat{j}+4\hat{k})

Comparing it with

\vec{r}=\vec{p}+t \vec{q},

we get

\vec{q}=(-\hat{i}-2\hat{j}+4\hat{k})

since,

\vec{q}\cdot \vec{n}=(-\hat{i}-2\hat{j}+4\hat{k})\cdot(42\hat{i}-3\hat{j}+9\hat{k})

=-42+6+36=0

Hence, the line given in option (a) is parallel to the given plane.

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