3d planes practice · Mixed difficulty

Option A is the correct answer because it represents the equation of a line passing through the given point and perpendicular to the line of intersection of the given planes.

Option C (-6) is correct because it is found by using the formula for the mirror image of a point in a plane and then substituting the values into the expression 2a+b+2c.

Option D (6) is correct because it represents the distance calculated by finding the intersection point of the line parallel to the z-axis and the plane, then determining the z-coordinate difference.

Option D ($\mathrm{k}=0$ or -3) is correct because these are the values of k that make the determinant of the matrix formed by the direction ratios and difference of points equal to zero, indicating coplanarity.

Option D (-5) is correct because it is the integer value of k that makes the two given lines intersect.

Option C (-16) is correct because substituting the points into the plane equation allows us to solve for a and b, and then calculate 3a + 5b.

Option D (5151) is correct because it's the result of calculating $\frac{N(N+1)}{2}$ where N is derived from the distance between the two planes.

Option A ($\frac{{p^5}}{{2 h k l}}$) is correct because it represents the area of the triangle ABC, derived from the equation of the plane and the intercepts on the coordinate axes.

Option A ($\left( {\sqrt 2 ,\,1,\,4} \right)$) is correct because, after finding the equation of the plane using the given points and angle condition, this point satisfies the equation.

Option C (c=0) is correct because for the plane to be perpendicular to the xy-plane, its normal vector must be orthogonal to the normal vector of the xy-plane, which is $\hat{k}$.

Option C ($3 x^2-10 x+7=0$) is correct because it is the equation satisfied by $\lambda$ when the points are equidistant from the plane.

Option A is correct because it represents the equations of the planes that satisfy the given conditions of passing through the line of intersection and being at a distance of 1 from the origin.

Option B (r. $(4 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=32$) is correct because it represents a plane parallel to the given plane and passing through the specified point.

Option C ((0,-3,1)) is correct because it satisfies the equation of the plane containing the given point and line.

Option A ($2 x-y-z=1$) is correct because it represents the plane formed by the intersection of the two given spheres.

Option A ($\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$) is correct because it is the foot of the perpendicular from P to the plane, calculated using the distance formula and the equation of the line perpendicular to the plane.

Option B ($x^2-18 x+72=0$) is correct because it involves finding the point of intersection of a line and a plane, then finding the distance of that point from another plane, and finally forming a quadratic equation with the given roots.

Option B (2) is correct. Since the point (-1, 2) lies outside the circle, there are two tangents that can be drawn from the point to the circle.

Option B (0) is correct because the point (1,2,-3) lies on the plane, making the shortest distance 0.

Option D ($4x - y - 5z + 2 = 0\,\,$) is the correct equation of the plane because it satisfies both the intersection condition and the distance from the origin.

Option A (5) is correct because the projection onto the xy-plane is the magnitude of the vector's x and y components, which is $\sqrt{3^2 + 4^2} = 5$.

Option B (2, -10, -2) is correct because by finding the value of $\alpha$ for which the lines are coplanar and substituting it into the equation of line $L_2$, the point (2, -10, -2) satisfies the equation.

Option A (10) is correct because it is the sum of the coordinates of the image point after reflection in the plane.

Option D (1) is correct because the distance is calculated by finding the intersection point of the line and plane, then finding the distance between the given point and the intersection point.

The correct answer is B. The shortest distance between the Z-axis and the given line is found by finding a plane containing the line that is parallel to the Z-axis, and then calculating the distance from a point on the Z-axis to that plane, which results in 2.

Option B ($\frac{2 a}{\sqrt{6}}$) is correct because it represents the shortest distance between the specified opposite edges of the tetrahedron.

Option B ($3x+2z=6$) is correct because it represents a plane parallel to the y-axis with the given intercepts on the x and z axes.

Option A (23) is correct because by finding the intersection point P of the line and the plane, and the intersection point Q of the line and the YZ-plane, the distance PQ can be calculated and expressed in the form a$\sqrt{b}$, leading to a+b = 23.

Option A ($20 x+23 y+26 z-69=0$) is correct because it represents the equation of a plane passing through the intersection of two given planes and a specific point.

Option C ($\frac{{x-1}}{{1}}=\frac{{y-1}}{{-2}}=\frac{{z+1}}{{-3}}$) is correct because the direction ratios of the line are the same as the direction ratios of the normal to the plane.