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1

### JEE Advanced 2021 Paper 1 Online

Numerical
Let $$\overrightarrow u$$, $$\overrightarrow v$$ and $$\overrightarrow w$$ be vectors in three-dimensional space, where $$\overrightarrow u$$ and $$\overrightarrow v$$ are unit vectors which are not perpendicular to each other and $$\overrightarrow u$$ . $$\overrightarrow w$$ = 1, $$\overrightarrow v$$ . $$\overrightarrow w$$ = 1, $$\overrightarrow w$$ . $$\overrightarrow w$$ = 4

If the volume of the paralleopiped, whose adjacent sides are represented by the vectors, $$\overrightarrow u$$, $$\overrightarrow v$$ and $$\overrightarrow w$$, is $$\sqrt 2$$, then the value of $$\left| {3\overrightarrow u + 5\overrightarrow v } \right|$$ is ___________.

2

### JEE Advanced 2021 Paper 1 Online

Numerical
Let $$\alpha$$, $$\beta$$ and $$\gamma$$ be real numbers such that the system of linear equations

x + 2y + 3z = $$\alpha$$

4x + 5y + 6z = $$\beta$$

7x + 8y + 9z = $$\beta$$ $$-$$ 1

is consistent. Let | M | represent the determinant of the matrix

$$M = \left[ {\matrix{ \alpha & \beta & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right]$$

Let P be the plane containing all those ($$\alpha$$, $$\beta$$, $$\gamma$$) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.

The value of D is _________.

## Explanation

$$7x + 8y + 9z - (\gamma - 1) = A(4x + 5y + 6z - \beta ) + B(x + 2y + 3z - \alpha )$$

On equating the coefficients,

4A + B = 7 .... (i)

5A + 2B = 8 .... (ii)

and $$-$$ ($$\gamma$$ $$-$$ 1) = $$-$$ A$$\beta$$ $$-$$ $$\alpha$$B ..... (iii)

On solving Eqs. (i) and (ii), we get A = 2 and B = $$-$$1

From Eq. (iii), we get

$$-$$ $$\gamma$$ + 1 = $$-$$ 2$$\beta$$ $$-$$ $$\alpha$$($$-$$1)

$$\Rightarrow$$ $$\alpha$$ $$-$$ 2$$\beta$$ + $$\gamma$$ = 1 ..... (iv)

Now, determinant of

$$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$$ [from Eq. (iv)]

Equation of plane P is given by $$x - 2y + z = 1$$

Hence, perpendicular distance of the point (0, 1, 0) from the plane

$$P = {{\left| {0 - 2 \times 1 + 0 - 1} \right|} \over {\sqrt {{1^2} + {{( - 2)}^2} + {1^2}} }} = {{\left| 3 \right|} \over {\sqrt 6 }}$$

$$\Rightarrow D = {\left( {{{\left| 3 \right|} \over {\sqrt 6 }}} \right)^2} = {9 \over 6} = 1.5$$
3

### JEE Advanced 2021 Paper 1 Online

Numerical
Let $$\alpha$$, $$\beta$$ and $$\gamma$$ be real numbers such that the system of linear equations

x + 2y + 3z = $$\alpha$$

4x + 5y + 6z = $$\beta$$

7x + 8y + 9z = $$\beta$$ $$-$$ 1

is consistent. Let | M | represent the determinant of the matrix

$$M = \left[ {\matrix{ \alpha & \beta & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right]$$

Let P be the plane containing all those ($$\alpha$$, $$\beta$$, $$\gamma$$) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.

The value of | M | is _________.

## Explanation

$$7x + 8y + 9z - (\gamma - 1) = A(4x + 5y + 6z - \beta ) + B(x + 2y + 3z - \alpha )$$

On equating the coefficients,

4A + B = 7 .... (i)

5A + 2B = 8 .... (ii)

and $$-$$ ($$\gamma$$ $$-$$ 1) = $$-$$ A$$\beta$$ $$-$$ $$\alpha$$B ..... (iii)

On solving Eqs. (i) and (ii), we get A = 2 and B = $$-$$1

From Eq. (iii), we get

$$-$$ $$\gamma$$ + 1 = $$-$$ 2$$\beta$$ $$-$$ $$\alpha$$($$-$$1)

$$\Rightarrow$$ $$\alpha$$ $$-$$ 2$$\beta$$ + $$\gamma$$ = 1 ..... (iv)

Now, determinant of

$$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$$ [from Eq. (iv)]
4

### JEE Advanced 2019 Paper 2 Offline

Numerical
Let $$\overrightarrow a = 2\widehat i + \widehat j - \widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j + \widehat k$$ be two vectors. Consider a vector c = $$\alpha$$$$\overrightarrow a$$ + $$\beta$$$$\overrightarrow b$$, $$\alpha$$, $$\beta$$ $$\in$$ R. If the projection of $$\overrightarrow c$$ on the vector ($$\overrightarrow a$$ + $$\overrightarrow b$$) is $$3\sqrt 2$$, then the
minimum value of ($$\overrightarrow c$$ $$-$$($$\overrightarrow a$$ $$\times$$ $$\overrightarrow b$$)).$$\overrightarrow c$$ equals ................

## Explanation

Given vectors $$\overrightarrow a$$$$= 2\widehat i + \widehat j - \widehat k$$

and $$\overrightarrow b = \widehat i + 2\widehat j + \widehat k$$

So, $$\overrightarrow a + \overrightarrow b = 3\widehat i + 3\widehat j \Rightarrow |\overrightarrow a + \overrightarrow b| = 3\sqrt 2$$

Since, it is given that projection of $$\overrightarrow c$$ = $$\alpha$$a + $$\beta$$b on the vector ($$\overrightarrow a$$ + $$\overrightarrow b$$) is $$3\sqrt 2$$, then

$${{(\overrightarrow a + \overrightarrow b ).\overrightarrow c } \over {|\overrightarrow a + \overrightarrow b|}} = 3\sqrt 2$$

$$\Rightarrow (\overrightarrow a + \overrightarrow b).(\alpha \overrightarrow a + \beta \overrightarrow b) = 18$$

$$\Rightarrow \alpha (\overrightarrow a.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) + \alpha (\overrightarrow b.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) = 18$$

$$\Rightarrow 6\alpha + 3\beta + 3\alpha + 6\beta = 18$$

$$\Rightarrow 9\alpha + 9\beta = 18 \Rightarrow (\alpha + \beta ) = 2$$ .....(i)

Now, for minimum value of

($$\overrightarrow c$$ $$-$$($$\overrightarrow a$$ $$\times$$ $$\overrightarrow b$$)).$$\overrightarrow c$$

$$= (\alpha \overrightarrow a + \beta \overrightarrow b - (\overrightarrow a \times \overrightarrow b)).(\alpha \overrightarrow a + \beta \overrightarrow b)$$

$$= {\alpha ^2}(\overrightarrow a.\overrightarrow a) + \alpha \beta (\overrightarrow a.\overrightarrow b) + \alpha \beta (\overrightarrow a.\overrightarrow b) + {\beta ^2}(\overrightarrow b.\overrightarrow b)$$

[$$\because$$ ($$\overrightarrow a$$ $$\times$$ $$\overrightarrow b$$) . $$\overrightarrow a$$ = 0 = ($$\overrightarrow a$$$$\times$$ $$\overrightarrow b$$) . $$\overrightarrow b$$]

$$6{\alpha ^2} + 6\alpha \beta + 6{\beta ^2} = 6({\alpha ^2} + {\beta ^2} + \alpha \beta )$$

$$= 6\,[{(\alpha + \beta )^2} - \alpha \beta ] = 6\,[4 - \alpha \beta ]$$

$$= 6\,[4 - \alpha (2 - \alpha )]$$

$$= 6\,[4 - 2\alpha + {\alpha ^2}]$$

Let f ( $$\alpha$$) = $$4 - 2\alpha + {\alpha ^2}$$

f′( $$\alpha$$) = –2 + 2$$\alpha$$

At maximum and minimum f′( $$\alpha$$) = 0

$$\Rightarrow$$ –2 + 2$$\alpha$$ $$\Rightarrow$$ $$\alpha$$ = 1

f′'( $$\alpha$$) = 2 (+ve)

Therefore, minimum value of $$4 - 2\alpha + {\alpha ^2}$$ is (4 – 2 + 1) = 3.

$$\therefore$$ The minimum value of

$$6(4 - 2\alpha + {\alpha^2}) = 6(3) = 18$$

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