Scalar triple product practice · Mixed difficulty

Option D ($x - 2y + z = 0$) is correct because it represents the equation of the plane containing the given line and perpendicular to the plane containing the other two lines, obtained by finding the normal vector and using the point-normal form.

Option A (neither x nor y) is correct because the scalar triple product [\vec{a} \vec{b} \vec{c}] simplifies to a constant value of 1, which is independent of both x and y.

Option A (5) is correct because the volume of the parallelepiped is found by calculating the absolute value of the scalar triple product of the given vectors.

The correct answer is C (3) because it is obtained by calculating the scalar triple product and using vector identities to simplify the expression for alpha.

Option A (${V^3}$) is correct because the volume of the new parallelopiped is the cube of the original volume, derived from determinant properties.

Option A is the correct answer because it is the only option that is not possible given the constraints of the problem and the calculated value of lambda.

Option C (no value of $\lambda$) is correct because the equation $(\lambda^2+1)\vec{a} \vec{b} \vec{c}=0$ has no real solution for $\lambda$ since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar.

Option A (-$\frac{146}{17}$) is correct because the points lie on a plane, so the volume of the tetrahedron formed by them is zero, leading to the given value of lambda.

Option D ($\frac{4}{3 \sqrt{3}}$) is correct because the volume of the parallelopiped formed by the unit vectors perpendicular to the planes is calculated using scalar triple product and vector properties.

Option D (9) is correct because it involves finding the shortest distance between two lines using the formula involving the cross product of direction vectors and the difference of position vectors.

Option B (-144) is correct because the scalar triple product a.(b x c) is calculated by the determinant of the matrix formed by the components of the vectors, which results in -144.

The correct answer is D ($\vec{r} \cdot(4 \hat{i}+3 \hat{j}+5 \hat{k})=50$) because the plane passes through the intersection point (4,3,5) and is perpendicular to the vector from the origin to that point, maximizing the distance from the origin.

Option A (3) is correct because the expression simplifies to the sum of the dot products of unit vectors with themselves, which equals 3.

Option A ($2\left( \mathbf{\vec a}\,\; \times \\,\mathbf{\vec b} \right)\,$) is correct because the expression simplifies to twice the cross product of the vectors $\vec{a}$ and $\vec{b}$.

Option B ($\frac{1}{\sqrt{2}}$) is correct because it is calculated by finding the volume of the tetrahedron and the area of the base, then using the formula for the height.

Option C (0) is correct because the condition r is perpendicular to (a + b + c) leads to l + m + n = 0.

Option A (0) is correct because the scalar triple product of coplanar vectors is zero.

Option D ($((P \wedge (\sim R)) \wedge Q) \vee ((\sim Q) \wedge ((\sim P) \wedge R))$) is correct because it represents the negation of the given statement using logical equivalences.

Option B ($\sqrt{35}$) is correct because the maximum value of the scalar triple product is the magnitude of the cross product of V and W.

Option C (both X-axis and Y-axis) is incorrect. Option D (both Y-axis and Z-axis) is correct because the line of intersection is perpendicular to the i vector, hence parallel to the YZ plane.

Option A (-2) is correct because it is the scalar triple product $a \cdot (b \times c)$, which can be calculated using the given vector equations and properties of cross and dot products.

Option D (\beginarraylll a⃗ & b⃗ & c⃗ \endarray²$\begin{{array}}{{lll}} \vec{{a}} & \vec{{b}} & \vec{{c}} \end{{array}}^2$) is correct because the scalar triple product identity allows the given expression to be simplified to the square of the scalar triple product of $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Option A (0) is correct because after calculating the vector triple product and comparing coefficients, the sum of lambda and mu is found to be 0.

Option D ($\sqrt {ab}$) is correct because coplanarity implies the scalar triple product is zero, leading to the relationship $c = \sqrt{ab}$.

Option C (0) is correct because the vectors are coplanar, making the scalar triple product equal to zero.

Option A ($\frac{\vec{{b}}}{{|\vec{{b}}|^2}}+\frac{{\vec{{a}} \times \vec{{b}}}}{{|\vec{{a}} \times \vec{{b}}|^2}}$) is correct because it satisfies the given conditions for the vector $\vec{{v}}$ when $\vec{{a}}$ and $\vec{{b}}$ are perpendicular.

Option D (8) is correct because the volume of the parallelepiped is calculated using the scalar triple product of the given vectors, which equals 8.

Option C ($\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]$) is correct because after expanding the scalar triple product, the result simplifies to the negative of the scalar triple product of a, b, and c.

Option D (- 215) is correct because the scalar triple product is evaluated using the properties of mutually perpendicular vectors.

Option B ($4\widehat {\\,\mathbf{\dot i}} + 5\widehat {\\,\mathbf{\dot j}} + 11\mathbf{\hat k}$) is correct because it is the cross product of the normal vectors of the two planes, which gives the direction vector of the line of intersection.