Let O be the origin. Let \overrightarrow OP = x\widehat i + y\widehat j - \widehat k and \overrightarrow OQ = - \widehat i + 2\widehat j + 3x\widehat k , x, y \in R, x > 0, be such that | \overrightarrow PQ | = \sqrt 20 and the vector \overrightarrow OP is perpendicular \overrightarrow OQ . If \overrightarrow OR = 3\widehat i + z\widehat j - 7\widehat k , z \in R, is coplanar with \overrightarrow OP and \overrightarrow OQ , then the value of x2 + y2 + z2 is equal to :

Vector Algebra PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q27 with Solution

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Correct answer: B

Write the given vectors

$\overrightarrow{OP}=(x,y,-1), \qquad \overrightarrow{OQ}=(-1,2,3x)$

Given:

$|\overrightarrow{PQ}|=\sqrt{20}$
$\overrightarrow{OP} \perp \overrightarrow{OQ}$
$\overrightarrow{OR}=(3,z,-7)$ is coplanar with $\overrightarrow{OP},\overrightarrow{OQ}$ .

We need $x^2+y^2+z^2$ .

Use perpendicularity condition

Since $\overrightarrow{OP} \cdot \overrightarrow{OQ}=0$ ,

$x(-1)+y(2)+(-1)(3x)=0$ $-x+2y-3x=0$ $2y-4x=0$ $y=2x$

Use the length of $\overrightarrow{PQ}$

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$

So,

$\overrightarrow{PQ}=(-1-x,\ 2-y,\ 3x+1)$

Given $|\overrightarrow{PQ}|=\sqrt{20}$ , hence

$(-1-x)^2+(2-y)^2+(3x+1)^2=20$

Using $y=2x$ ,

$(-1-x)^2+(2-2x)^2+(3x+1)^2=20$

Now expand:

$(x+1)^2 + 4(1-x)^2 + (3x+1)^2 = 20$

$(x^2+2x+1) + 4(x^2-2x+1) + (9x^2+6x+1)=20$

$x^2+2x+1+4x^2-8x+4+9x^2+6x+1=20$

$14x^2+0x+6=20$

$14x^2=14$ $x^2=1$

Since $x>0$ , we get

$x=1$

Then

$y=2x=2$

Use coplanarity of $\overrightarrow{OR}$ with $\overrightarrow{OP},\overrightarrow{OQ}$

Now,

$\overrightarrow{OP}=(1,2,-1), \qquad \overrightarrow{OQ}=(-1,2,3), \qquad \overrightarrow{OR}=(3,z,-7)$

For coplanarity, scalar triple product must be zero:

1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7 \end{vmatrix}=0$$ Expand along first row: $$1\begin{vmatrix}2 & 3 \\ z & -7\end{vmatrix} -2\begin{vmatrix}-1 & 3 \\ 3 & -7\end{vmatrix} +(-1)\begin{vmatrix}-1 & 2 \\ 3 & z\end{vmatrix}=0$$ $$1(2\cdot(-7)-3z)-2[(-1)(-7)-3\cdot3]-[(-1)z-2\cdot3]=0$$ $$(-14-3z)-2(7-9)-(-z-6)=0$$ $$(-14-3z)-2(-2)+z+6=0$$ $$-14-3z+4+z+6=0$$ $$-4-2z=0$$ $$z=-2$$ --- 5. **Compute required value** $$x^2+y^2+z^2=1^2+2^2+(-2)^2=1+4+4=9$$ --- 6. **Compare with stored answer** Derived answer is **9**, which matches option **B** and agrees with the stored correct answer.

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View written solutionFree

Correct answer: B

Write the given vectors

$\overrightarrow{OP}=(x,y,-1), \qquad \overrightarrow{OQ}=(-1,2,3x)$

Given:

$|\overrightarrow{PQ}|=\sqrt{20}$
$\overrightarrow{OP} \perp \overrightarrow{OQ}$
$\overrightarrow{OR}=(3,z,-7)$ is coplanar with $\overrightarrow{OP},\overrightarrow{OQ}$ .

We need $x^2+y^2+z^2$ .

Use perpendicularity condition

Since $\overrightarrow{OP} \cdot \overrightarrow{OQ}=0$ ,

$x(-1)+y(2)+(-1)(3x)=0$ $-x+2y-3x=0$ $2y-4x=0$ $y=2x$

Use the length of $\overrightarrow{PQ}$

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$

So,

$\overrightarrow{PQ}=(-1-x,\ 2-y,\ 3x+1)$

Given $|\overrightarrow{PQ}|=\sqrt{20}$ , hence

$(-1-x)^2+(2-y)^2+(3x+1)^2=20$

Using $y=2x$ ,

$(-1-x)^2+(2-2x)^2+(3x+1)^2=20$

Now expand:

$(x+1)^2 + 4(1-x)^2 + (3x+1)^2 = 20$

$(x^2+2x+1) + 4(x^2-2x+1) + (9x^2+6x+1)=20$

$x^2+2x+1+4x^2-8x+4+9x^2+6x+1=20$

$14x^2+0x+6=20$

$14x^2=14$ $x^2=1$

Since $x>0$ , we get

$x=1$

Then

$y=2x=2$

Use coplanarity of $\overrightarrow{OR}$ with $\overrightarrow{OP},\overrightarrow{OQ}$

Now,

$\overrightarrow{OP}=(1,2,-1), \qquad \overrightarrow{OQ}=(-1,2,3), \qquad \overrightarrow{OR}=(3,z,-7)$

For coplanarity, scalar triple product must be zero:

1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7 \end{vmatrix}=0$$ Expand along first row: $$1\begin{vmatrix}2 & 3 \\ z & -7\end{vmatrix} -2\begin{vmatrix}-1 & 3 \\ 3 & -7\end{vmatrix} +(-1)\begin{vmatrix}-1 & 2 \\ 3 & z\end{vmatrix}=0$$ $$1(2\cdot(-7)-3z)-2[(-1)(-7)-3\cdot3]-[(-1)z-2\cdot3]=0$$ $$(-14-3z)-2(7-9)-(-z-6)=0$$ $$(-14-3z)-2(-2)+z+6=0$$ $$-14-3z+4+z+6=0$$ $$-4-2z=0$$ $$z=-2$$ --- 5. **Compute required value** $$x^2+y^2+z^2=1^2+2^2+(-2)^2=1+4+4=9$$ --- 6. **Compare with stored answer** Derived answer is **9**, which matches option **B** and agrees with the stored correct answer.

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