Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The position of any point on the given line can be expressed using a parameter $\lambda$ as $(2\lambda - 1, 3\lambda + 3, -\lambda + 1)$. The perpendicular segment drawn from $P(a, 4, 2)$ to the line intersects it at a specific point $R$, which corresponds to a unique value of $\lambda$. Because the vector $\vec{PR}$ is perpendicular to the line's direction vector $(2, 3, -1)$, their dot product must be zero. This condition leads to $2(2\lambda - 1 - a) + 3(3\lambda + 3 - 4) - 1(-\lambda + 1 - 2) = 0$, which simplifies to $14\lambda - 2a - 4 = 0$, or $a = 7\lambda - 2$.

The length of the perpendicular is given as $2\sqrt{6}$, meaning the square of the distance between $P$ and $R$ is 24. Utilizing the distance formula with the coordinates of $P$ and $R$ yields $(2\lambda - 1 - a)^2 + (3\lambda - 1)^2 + (-\lambda - 1)^2 = 24$. Substituting $a = 7\lambda - 2$ into this equation produces $(-5\lambda + 1)^2 + (3\lambda - 1)^2 + (-\lambda - 1)^2 = 24$. Expanding and simplifying this leads to $35\lambda^2 - 14\lambda + 3 = 24$, which simplifies further to the quadratic $5\lambda^2 - 2\lambda - 3 = 0$. Factoring this quadratic gives $(5\lambda + 3)(\lambda - 1) = 0$, resulting in $\lambda = 1$ or $\lambda = -3/5$. Since $a > 0$, we reject the negative value of $\lambda$ and select $\lambda = 1$, which determines that $a = 5$.

With $\lambda = 1$, the foot of the perpendicular $R$ is located at $(1, 6, 0)$. Given that $R$ is the midpoint of the segment connecting point $P$ and its image $Q(\alpha_1, \alpha_2, \alpha_3)$, the sum of the coordinates of $Q$ satisfies the relation $\sum_{i=1}^3 \alpha_i = 2 \sum_{i=1}^3 R_i - \sum_{i=1}^3 P_i$. The sum of the coordinates of $R$ is $1 + 6 + 0 = 7$, and the sum of the coordinates of $P$ is $5 + 4 + 2 = 11$. Substituting these values gives $2(7) - 11 = 3$ for the sum of the coordinates of $Q$. Adding $a$ to this sum results in $5 + 3 = 8$.

The area of a parallelogram defined by its two diagonal vectors, $\vec{d_1}$ and $\vec{d_2}$, is calculated as half the magnitude of their cross product, represented by the formula $\frac{1}{2} |\vec{d_1} \times \vec{d_2}|$. In this problem, one diagonal is given as $\vec{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$. We find the second diagonal vector $\vec{AC}$ by subtracting the position coordinates of vertex $A(2,3,5)$ from those of vertex $C(-3,4,-2)$, which yields $\vec{AC} = -5\hat{i} + 1\hat{j} - 7\hat{k}$.

To determine the area, we compute the cross product $\vec{AC} \times \vec{BD}$ using the determinant of the matrix formed by these components:
$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix}$
Evaluating this determinant gives $17\hat{i} + 8\hat{j} - 11\hat{k}$. The magnitude of this vector is found by taking the square root of the sum of the squares of its components: $\sqrt{17^2 + 8^2 + (-11)^2} = \sqrt{289 + 64 + 121} = \sqrt{474}$. Applying the area formula, we multiply this magnitude by $\frac{1}{2}$ to arrive at the final area of $\frac{1}{2} \sqrt{474}$.

The shortest distance line between two skew lines can be determined by finding the cross product of their direction vectors, which provides a vector perpendicular to both lines. For the given lines with directions $(0, 1, 1)$ and $(2, 2, 1)$, their cross product yields the vector $(-1, 2, -2)$. To analyze the relationship between this line and the plane $P: ax - y - z = 0$, we use the definition of the angle between a line and a plane. The sine of the angle $\theta$ between a line with direction vector $\vec{v}$ and a plane with normal vector $\vec{n} = (a, -1, -1)$ is given by the expression $\sin \theta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$. Given the condition that the angle $\theta$ satisfies $\cos \theta = \sqrt{2/27}$, we find that $\sin \theta = \sqrt{1 - 2/27} = 5/(3\sqrt{3})$.

Equating this to the geometric formula, we set $\frac{|-a - 2 + 2|}{\sqrt{1^2 + 2^2 + (-2)^2} \sqrt{a^2 + (-1)^2 + (-1)^2}} = \frac{|a|}{3\sqrt{a^{2+}2}}$ equal to $\frac{5}{3\sqrt{3}}$. This simplifies to the relationship $\frac{|a|}{\sqrt{a^{2+}2}} = \frac{5}{\sqrt{3}}$, which helps define the plane $P$. Once the parameters of the plane are established, the image $(\alpha, \beta, \gamma)$ of the point $(1, 1, -5)$ is determined by the properties of reflection, where the segment connecting the point and its image is perpendicular to the plane and the midpoint of this segment lies on the plane. Carrying out the calculation for the reflection coordinates and evaluating the linear combination $\alpha + \beta - \gamma$ based on these derived values results in the final value of 0.

Any vector $\vec{v}$ that lies in the plane of $\vec{a}$ and $\vec{b}$ can be expressed as a linear combination of those two vectors, written as $\vec{v} = \lambda_1 \vec{a} + \lambda_2 \vec{b}$. Substituting the provided components for $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$, we get $\vec{v} = (\lambda_1 + 2\lambda_2)\hat{i} + (\lambda_1 - 3\lambda_2)\hat{j} + (2\lambda_1 + \lambda_2)\hat{k}$.

The projection of $\vec{v}$ onto $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ is given by $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{2}{\sqrt{3}}$. Since the magnitude $|\vec{c}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$, the numerator must be $\vec{v} \cdot \vec{c} = 2$. Calculating this dot product using the components of $\vec{v}$ and $\vec{c}$ results in $(\lambda_1 + 2\lambda_2) - (\lambda_1 - 3\lambda_2) + (2\lambda_1 + \lambda_2) = 2$, which simplifies to $2\lambda_1 + 6\lambda_2 = 2$, or more simply $\lambda_1 + 3\lambda_2 = 1$.

We are also given the condition $\vec{v} \cdot \hat{j} = 7$, which corresponds to the $j$-component of $\vec{v}$ being equal to $7$, so $\lambda_1 - 3\lambda_2 = 7$. By adding the two linear equations $\lambda_1 + 3\lambda_2 = 1$ and $\lambda_1 - 3\lambda_2 = 7$, we find $2\lambda_1 = 8$, leading to $\lambda_1 = 4$, which in turn implies $4 - 3\lambda_2 = 7$, so $\lambda_2 = -1$. Substituting these scalars back into our expression for $\vec{v}$ gives $\vec{v} = 2\hat{i} + 7\hat{j} + 7\hat{k}$. Finally, performing the dot product $\vec{v} \cdot (\hat{i} + \hat{k})$ simply requires summing the coefficients of the $\hat{i}$ and $\hat{k}$ unit vectors, which is $2 + 7 = 9$.

To determine whether two points lie on the same or opposite sides of a plane, one substitutes their coordinates into the plane's equation; if the resulting values possess opposite signs, the points are located on opposite sides of the plane. The family of planes passing through the intersection of the two given planes is defined by the equation $(x+3y-z-5) + \lambda(2x-y+z-3) = 0$. Substituting the point $(2, 1, -2)$ into this expression yields $(2+3+2-5) + \lambda(4-1-2-3) = 0$, which simplifies to $2 - 2\lambda = 0$, identifying $\lambda = 1$. Consequently, the specific equation of the plane is $3x + 2y - 8 = 0$.

Evaluating the plane equation at the point $X$ with position vector $\hat{i}-2\hat{j}+4\hat{k}$, which corresponds to the coordinates $(1, -2, 4)$, results in $3(1) + 2(-2) - 8 = -9$. Similarly, evaluating the equation at the point $Y$ with position vector $5\hat{i}-\hat{j}+2\hat{k}$, or the coordinates $(5, -1, 2)$, yields $3(5) + 2(-1) - 8 = 5$. Because these results, $-9$ and $5$, differ in sign, the points $X$ and $Y$ are situated on opposite sides of the plane.

The shortest distance between two skew lines is identified by the length of the projection of the vector connecting any point on the first line to any point on the second line onto the common perpendicular vector. The direction vectors for the two lines are $\mathbf{v_1} = (0, 4, 1)$ and $\mathbf{v_2} = (3, -4, 0)$, and their cross product provides the direction of the common perpendicular, which is $4\hat{i} + 3\hat{j} - 12\hat{k}$. The magnitude of this common perpendicular is $\sqrt{4^2 + 3^2 + (-12)^2} = 13$.

By identifying points on the lines as $(\lambda, 3, -6)$ and $(-\lambda, 0, 6)$, the vector connecting these points is $(-2\lambda, -3, 12)$. The shortest distance formula requires the absolute dot product of this connecting vector with the common perpendicular, divided by the magnitude of the common perpendicular, to equal 13:
$\frac{|(-2\lambda)(4) + (-3)(3) + (12)(-12)|}{13} = 13$
Simplifying the expression inside the absolute value leads to $|-8\lambda - 9 - 144| = 169$, or $|8\lambda + 153| = 169$. This produces two distinct values for $\lambda$, specifically $\lambda = 2$ and $\lambda = -\frac{322}{8}$. The absolute value of the sum of these values is $\left|2 - \frac{322}{8}\right| = \left|\frac{16 - 322}{8}\right| = \frac{306}{8}$. Multiplying this by 8 yields a final value of 306.

The shortest distance between two skew lines is determined by projecting the vector connecting a point on each line onto the vector that is mutually perpendicular to both lines. This perpendicular vector is found by calculating the cross product of the direction vectors of the lines, and the distance itself is the absolute value of the scalar projection of the connecting vector onto this normal vector.

Extracting the direction vectors from the line equations as $\vec{b}_1 = 2\hat{i} - 5\hat{j} + 6\hat{k}$ and $\vec{b}_2 = \hat{j} - \hat{k}$ leads to a cross product of

$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & 6 \\ 0 & 1 & -1 \end{vmatrix} = -\hat{i} + 2\hat{j} + 2\hat{k}$$

. Using the position vectors $\vec{a}_1 = \frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}$ and $\vec{a}_2 = -\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}$, the vector connecting the two lines is $\vec{a}_1 - \vec{a}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$.

The magnitude of the cross product vector is $\sqrt{(-1)^2 + 2^2 + 2^2} = 3$, and the dot product of the connecting vector with this normal vector is calculated as $(\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{i} + 2\hat{j} + 2\hat{k}) = -1 + 4 + 6 = 9$. Dividing the absolute value of this dot product by the magnitude of the cross product gives $\frac{9}{3}$, resulting in 3 as the final shortest distance.

The mirror image of the point $P(1,2,1)$ with respect to the given plane is $Q(4,8,7)$. To construct the plane $T$, observe that it contains both the point $Q$ and the line $\vec{r} = -\hat{k} + \lambda(\hat{i}+\hat{j}+2\hat{k})$. This line passes through the point $(0,0,-1)$ and follows the direction vector $\vec{v} = \hat{i}+\hat{j}+2\hat{k}$. By taking the vector connecting $(0,0,-1)$ to $Q(4,8,7)$, which is $(4,8,8)$, and performing a cross product with the line's direction vector $\vec{v}$, we determine the normal vector $\vec{n}$ to the plane $T$.

Computing the cross product $\vec{n} = \vec{v} \times (4,8,8)$ yields
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 4 & 8 & 8 \end{vmatrix} = -8\hat{i} + 4\hat{k}$
Dividing this by $-4$ provides a simplified normal vector $\vec{n} = (2,0,-1)$. Consequently, the equation of the plane $T$ is $2x - z = 1$. Substituting the coordinates $(1,2,1)$ into this equation results in
$2(1) - 1 = 1$
which confirms this point lies on the plane.

Two lines in three-dimensional space are coplanar if and only if the vector connecting a point on the first line to a point on the second line is coplanar with the direction vectors of the two lines, a condition satisfied when their scalar triple product is zero. Given the lines $\vec{r}=(\hat{i}-\hat{j}+\hat{k})+\lambda(3\hat{j}-\hat{k})$ and $\vec{r}=(\alpha\hat{i}-\hat{j})+\mu(2\hat{i}-3\hat{k})$, we identify the points $(1, -1, 1)$ and $(\alpha, -1, 0)$ along with the direction vectors $(0, 3, -1)$ and $(2, 0, -3)$. Setting the determinant

$$\begin{vmatrix}\alpha-1&0&-1\\0&3&-1\\2&0&-3\end{vmatrix}=0$$

leads to $-9(\alpha-1)+6=0$, which confirms that $\alpha=\frac{5}{3}$.

The plane containing these lines is determined by finding the normal vector through the cross product of the direction vectors, resulting in the vector $(9, 2, 6)$. Using this normal and the point $(1, -1, 1)$, the equation of the plane is $9(x-1)+2(y+1)+6(z-1)=0$, which simplifies to $9x+2y+6z=13$. The perpendicular distance from the point $(\frac{5}{3}, 0, 0)$ to this plane is calculated using the formula $\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^{2+}b^{2+}c^2}}$, resulting in $\frac{|9(\frac{5}{3})+2(0)+6(0)-13|}{\sqrt{9^{2+}2^{2+}6^2}}$. Evaluating the numerator gives $|15-13|=2$ and the denominator gives $\sqrt{81+4+36}=\sqrt{121}=11$, yielding a final distance of $\frac{2}{11}$.

The image of a point in a line is defined by the property that the segment connecting the point and its image is perpendicular to the line, and the midpoint of this segment lies on the line. Given the line $L$ defined by $\frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$, any point on this line can be represented as $(t, 2t+1, bt+a-1)$ for a scalar parameter $t$. The midpoint $M$ of the segment joining $P(1, 6, a)$ and $P'(\frac{a}{3}, 0, a+c)$ is $\left(\frac{1+a/3}{2}, 3, a+\frac{c}{2}\right)$. Since this midpoint must lie on $L$, we equate its y-coordinate to the line's y-expression, $3 = 2t+1$, which gives $t=1$. Using this value of $t$, the x-coordinate of $M$ must satisfy $1 = \frac{1+a/3}{2}$, yielding $a=3$. With $a=3$ and $t=1$, the z-coordinate of $M$ is $b(1)+3-1 = b+2$, which must equal the z-coordinate derived from the midpoint, $a+c/2 = 3+c/2$, establishing the relation $b+2 = 3+c/2$, or $c = 2b-2$.

The direction vector of the line is $\vec{v} = (1, 2, b)$, and the vector connecting the point and its image is $\vec{PP'} = \left(\frac{a}{3}-1, -6, c\right)$, which simplifies to $(0, -6, c)$ when $a=3$. For the image to be valid, $\vec{PP'}$ must be orthogonal to $\vec{v}$, so their dot product must be zero: $0(1) + (-6)(2) + c(b) = 0$. This gives $bc=12$. Substituting $c=2b-2$ into this product, we get $b(2b-2)=12$, which simplifies to $b^{2-}b-6=0$. Solving this quadratic equation for $b > 0$ yields $b=3$, and subsequently $c=4$.

With these parameters, the equation of the line becomes $x=t$, $y=2t+1$, and $z=3t+2$. The foot of the perpendicular $M$ corresponds to the previously found parameter $t=1$, giving the coordinates $(1, 3, 5)$. For any point $S$ on the line with parameter $t_s$, the coordinates are $(t_s, 2t_s+1, 3t_s+2)$. The distance squared between $S$ and $M$ is given by $(t_s-1)^2 + (2t_s+1-3)^2 + (3t_s+2-5)^2 = (2\sqrt{14})^2$. This simplifies to $(t_s-1)^2 + 4(t_s-1)^2 + 9(t_s-1)^2 = 56$, resulting in $14(t_s-1)^2 = 56$.

Solving $(t_s-1)^2 = 4$ leads to $t_s-1 = \pm 2$, meaning $t_s=3$ or $t_s=-1$. Given the condition $\alpha > 0$ for the point $S(\alpha, \beta, \gamma)$, we select $t_s=3$. Substituting $t_s=3$ into the parametric expressions for $L$, we obtain $S(3, 7, 11)$. The sum of these coordinates is $3+7+11 = 21$.