JEE Mathematics 2023 Question with Solution

Given: A vector $\mathbf{v}$ in the first octant makes angles $60^\circ$, $45^\circ$ and $\gamma$ with the x, y and z axes respectively. A plane through $\left( \sqrt{2}, -1, 1 \right)$ and $(a,b,c)$ is normal to this vector.

Find: The correct relation among $a, b, c$.

For a vector with direction cosines,

$$\hat{v} = \cos 60^\circ \, \hat{i} + \cos 45^\circ \, \hat{j} + \cos \gamma \, \hat{k}$$

Using the identity of direction cosines,

$$\cos^2 60^\circ + \cos^2 45^\circ + \cos^2 \gamma = 1$$ $$\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$$

Since $\gamma$ is acute,

$$\cos \gamma = \frac{1}{2}$$

Hence $\gamma = 60^\circ$.

Therefore a normal vector to the plane is proportional to

$$\left( \cos 60^\circ, \cos 45^\circ, \cos 60^\circ \right) = \left( \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \right)$$

So the equation of the plane through $\left( \sqrt{2}, -1, 1 \right)$ is

$$\frac{1}{2}(x-\sqrt{2}) + \frac{1}{\sqrt{2}}(y+1) + \frac{1}{2}(z-1) = 0$$

Multiplying and simplifying,

$$x + \sqrt{2}y + z = 1$$

Since $(a,b,c)$ lies on this plane,

$$a + \sqrt{2}b + c = 1$$

Therefore the relation obtained from the working is $a + \sqrt{2}b + c = 1$.

The solution concludes "The Correct Option is B", but this disagrees with the listed options. In the given options, the matching relation is option C.

Given: The plane is normal to $\mathbf{v}$.

A plane normal is obtained directly from the direction ratios of $\mathbf{v}$. Since the vector lies in the first octant, all direction cosines are positive.

$$l = \cos 60^\circ = \frac{1}{2}, \quad m = \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad n = \cos \gamma$$

Now,

$$l^2 + m^2 + n^2 = 1$$ $$\frac{1}{4} + \frac{1}{2} + n^2 = 1$$ $$n^2 = \frac{1}{4}$$

Since the angle with the z-axis is acute, $n>0$, so

$$n = \frac{1}{2}$$

Thus the normal vector can be taken as

$$\left( \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \right)$$

Using point-normal form through $\left( \sqrt{2}, -1, 1 \right)$,

$$\frac{1}{2}(x-\sqrt{2}) + \frac{1}{\sqrt{2}}(y+1) + \frac{1}{2}(z-1) = 0$$

Multiply by $2$:

$$(x-\sqrt{2}) + \sqrt{2}(y+1) + (z-1) = 0$$ $$x - \sqrt{2} + \sqrt{2}y + \sqrt{2} + z - 1 = 0$$ $$x + \sqrt{2}y + z - 1 = 0$$

Hence,

$$x + \sqrt{2}y + z = 1$$

Substituting the point $(a,b,c)$,

$$a + \sqrt{2}b + c = 1$$

So the correct listed option is C.