Q.Show that the line through the points $(1, -1, 2), (3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If the line joining $A(4,1,2)$ and $B(0,k,1)$ is perpendicular to the line joining $C(-2,1,1)$ and $D(4,2,5)$, then the value of $k =$ ______ (A) $31$ (B) $-29$ (C) $-31$ (D) $29$

   ›Reveal solutionSolution

   Perpendicular lines have direction vectors with zero dot product; setting up $\vec{AB}\cdot\vec{CD}=0$ gives $k=29$.

   Concept and Intuition

   Two lines are perpendicular exactly when their direction vectors have a zero dot product. Here, $\vec{AB}$ is the direction of the line through $A,B$ and $\vec{CD}$ is the direction of the line through $C,D$.

   Step-by-Step Solution

   1. $\vec{AB}=B-A=(0-4,\,k-1,\,1-2)=(-4,\,k-1,\,-1)$.
   2. $\vec{CD}=D-C=(4-(-2),\,2-1,\,5-1)=(6,\,1,\,4)$.
   3. Perpendicularity: $\vec{AB}\cdot\vec{CD}=0$:

   $(-4)(6)+(k-1)(1)+(-1)(4)=0$

   $-24+k-1-4=0 \Rightarrow k-29=0 \Rightarrow k=29.$

   Common Mistakes

   • Computing $\vec{AB}$ as $A-B$ instead of $B-A$ (sign doesn't matter for the dot-product-zero condition, but consistency matters for other problems).
   • Arithmetic slip combining the constants $-24-1-4=-29$ vs correctly isolating $k$.

   ✓Final answer

   The correct option is (D) — $29$.

   ANSWER: D

2. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$, and is perpendicular to the plane $2x + 3y + z = 5$. (A) $3x - 4y + 18z + 32 = 0$ (B) $3x + 4y - 18z + 32 = 0$ (C) $4x + 3y - z + 1 = 0$ (D) $4x - 3y + z + 1 = 0$

   ›Reveal solutionSolution

   Three linear conditions (two points + perpendicularity to a given plane) on $ax+by+cz+d=0$ pin the plane down to $4x-3y+z+1=0$.

   Concept and Intuition

   A plane through two given points must satisfy each point's coordinates in its equation. "Perpendicular to another plane" means the two planes' normal vectors are perpendicular, i.e. their dot product is zero. Three such linear conditions (two point conditions + one perpendicularity condition) determine the plane's coefficients up to a common scale.

   Step-by-Step Solution

   1. Let the plane be $ax+by+cz+d=0$.
   2. Through $(0,1,2)$: $b+2c+d=0$ … (i)
   3. Through $(-1,0,3)$: $-a+3c+d=0$ … (ii)
   4. Perpendicular to $2x+3y+z=5$ (normal $(2,3,1)$): $2a+3b+c=0$ … (iii)
   5. From (i): $d=-b-2c$. From (ii): $d=a-3c$. Equate: $-b-2c=a-3c \Rightarrow c=a+b$.
   6. Substitute into (iii): $2a+3b+(a+b)=0 \Rightarrow 3a+4b=0 \Rightarrow a=-\dfrac{4b}{3}$.
   7. Choose $b=3$ (clears the fraction): $a=-4$, $c=a+b=-4+3=-1$, $d=-b-2c=-3-2(-1)=-1$.
   8. Plane: $-4x+3y-z-1=0$, i.e. multiplying by $-1$: $4x-3y+z+1=0$.
   9. Verify $(0,1,2)$: $0-3+2+1=0$. Verify $(-1,0,3)$: $-4-0+3+1=0$. Verify perpendicularity: $(4,-3,1)\cdot(2,3,1)=8-9+1=0$. All check out.

   Common Mistakes

   • Sign errors when equating the two expressions for $d$.
   • Forgetting that the two-point condition alone doesn't fully determine the plane — the perpendicularity condition is essential to fix the third degree of freedom.

   ✓Final answer

   The correct option is (D) — $4x-3y+z+1=0$.

   ANSWER: D

3. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.Let $u = 2i + 3j + k$, $v = -3i + 2j$ and $w = i - j + 4k$. Then which of the following statement is true? (A) $u$ is perpendicular to $v$ but not $w$ (B) $v$ is perpendicular to $w$ but not $u$ (C) $w$ is perpendicular to $u$ but not $v$ (D) $u$ is perpendicular to both $v$ and $w$

   ›Reveal solutionSolution

   Direct dot products show $u\cdot v=0$ (perpendicular) and $u\cdot w=3\ne0$ (not perpendicular), matching option (A).

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Checking each pair's dot product directly settles every option — no need for angle or cross-product computation.

   Step-by-Step Solution

   1. $u\cdot v = (2)(-3)+(3)(2)+(1)(0) = -6+6+0 = 0$ — so $u\perp v$.
   2. $u\cdot w = (2)(1)+(3)(-1)+(1)(4) = 2-3+4 = 3 \ne 0$ — so $u$ is not perpendicular to $w$.
   3. For completeness, $v\cdot w=(-3)(1)+(2)(-1)+(0)(4)=-3-2+0=-5\ne0$ — $v$ is also not perpendicular to $w$.
   4. So exactly the statement "$u$ is perpendicular to $v$ but not $w$" is true.

   Common Mistakes

   • Sign errors when multiplying negative components.
   • Checking only one dot product and assuming the others follow without verifying.

   ✓Final answer

   The correct option is (A) — $u$ is perpendicular to $v$ but not $w$.

   ANSWER: A

4. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The locus of a point at which the line joining the points $(-3, 1, 2), (1, -2, 4)$ subtends a right angle, is (A) $x^2 + y^2 + z^2 + 2x + y - 6z - 3 = 0$ (B) $x^2 + y^2 + z^2 + 2x - y - 6z + 3 = 0$ (C) $x^2 + y^2 + z^2 + 2x + y - 6z + 3 = 0$ (D) $x^2 + y^2 + z^2 - 2x + y - 6z + 3 = 0$

   ›Reveal solutionSolution

   This tests the classic 'locus subtending a right angle' problem, which is just the sphere having $AB$ as diameter, expressed via a perpendicularity dot-product condition. Answer: $x^2+y^2+z^2+2x+y-6z+3=0$.

   Concept and Intuition

   If a segment $AB$ subtends a right angle at a variable point $P$, then $PA\perp PB$, i.e. $\vec{PA}\cdot\vec{PB}=0$ for every such $P$ — this is exactly the defining property of a sphere with $AB$ as diameter (angle in a semicircle is a right angle, generalized to 3D). Writing this dot product in coordinates directly gives the sphere's equation.

   Step-by-Step Solution

   1. Let $P=(x,y,z)$. Then $\vec{PA}=A-P=(-3-x,\,1-y,\,2-z)$ and $\vec{PB}=B-P=(1-x,\,-2-y,\,4-z)$.
   2. Right angle at $P$: $\vec{PA}\cdot\vec{PB}=0$.
   3. $(-3-x)(1-x)=x^2+2x-3$ (expand: $-3+3x-x+x^2$).
   4. $(1-y)(-2-y)=y^2+y-2$ (expand: $-2-y+2y+y^2$).
   5. $(2-z)(4-z)=z^2-6z+8$ (expand: $8-2z-4z+z^2$).
   6. Sum to zero: $x^2+y^2+z^2+2x+y-6z+(-3-2+8)=0 \Rightarrow x^2+y^2+z^2+2x+y-6z+3=0$.

   Common Mistakes

   • Sign errors expanding each bracket pair — easiest to slip on the middle ($y$) term.
   • Using $\vec{AP}\cdot\vec{BP}$ vs $\vec{PA}\cdot\vec{PB}$ inconsistently; both give the same dot product since a sign flip in both factors cancels, so this isn't actually a pitfall here, but mixing one flipped and one not is.

   ✓Final answer

   The correct option is (C) — $x^2+y^2+z^2+2x+y-6z+3=0$.

   ANSWER: C

5. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the equation of the plane passing through the points $(1,-3,2)$, $(-2,3,1)$ and perpendicular to the plane $x + 2y - 3z = 0$ is $ax + by + cz + d = 0$, then $\dfrac{a+b}{c+d} =$ (A) $\dfrac{3}{11}$ (B) $13$ (C) $\dfrac{13}{11}$ (D) $3$

   ›Reveal solutionSolution

   This tests finding a plane through two points and perpendicular to another plane, using the cross product of the connecting direction vector and the given plane's normal. Answer: $13$.

   Concept and Intuition

   A plane's normal vector must be perpendicular to every direction lying in the plane. Since the plane contains points $P_1(1,-3,2)$ and $P_2(-2,3,1)$, the vector $\vec{P_1P_2}$ lies in the plane, so the required normal $\vec n=(a,b,c)$ satisfies $\vec n\cdot\vec{P_1P_2}=0$. Also, "perpendicular to the plane $x+2y-3z=0$" means the two planes' normals are perpendicular, so $\vec n\cdot(1,2,-3)=0$. A vector perpendicular to both $\vec{P_1P_2}$ and $(1,2,-3)$ is simply their cross product.

   Step-by-Step Solution

   1. Direction vector: $\vec d = P_2-P_1 = (-2-1,\,3-(-3),\,1-2)=(-3,6,-1)$.
   2. Normal of given plane: $\vec n_2=(1,2,-3)$.
   3. Required normal: $\vec n=\vec d\times \vec n_2 = \begin{vmatrix}\vec i&\vec j&\vec k\\-3&6&-1\\1&2&-3\end{vmatrix}$ $= \vec i(6(-3)-(-1)(2)) - \vec j((-3)(-3)-(-1)(1)) + \vec k((-3)(2)-6(1))$ $= \vec i(-18+2) - \vec j(9+1) + \vec k(-6-6) = (-16,-10,-12)$.
   4. Simplify (divide by $-2$): $\vec n \parallel (8,5,6)$, so $a=8,\,b=5,\,c=6$.
   5. Plane: $8x+5y+6z+d=0$. Substitute $(1,-3,2)$: $8-15+12+d=0 \Rightarrow d=-5$ (check with $(-2,3,1)$: $-16+15+6-5=0$ ✓).
   6. $\dfrac{a+b}{c+d}=\dfrac{8+5}{6+(-5)}=\dfrac{13}{1}=13$.

   Common Mistakes

   • Forgetting that "perpendicular to a plane" translates to a perpendicular condition on the normals, not the planes' points.
   • Sign errors while expanding the $3\times3$ determinant for the cross product.

   ✓Final answer

   The correct option is (B) — $13$.

   ANSWER: B

6. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The set of real values of λ for which the vectors $\lambda\vec{i}-3\vec{j}+5\vec{k}$ and $2\lambda\vec{i}-\lambda\vec{j}+\vec{k}$ are perpendicular to each other is (A) $\{0, 1\}$ (B) $\{-2\}$ (C) $\{2, -1\}$ (D) $\varphi$

   ›Reveal solutionSolution

   Perpendicular vectors have zero dot product; the resulting quadratic in $\lambda$ has no real roots, so the answer set is empty.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product vanishes. Setting up that equation converts a geometry condition into an algebraic one in $\lambda$.

   Step-by-Step Solution

   1. The vectors are $\vec u=(\lambda,-3,5)$ and $\vec v=(2\lambda,-\lambda,1)$.
   2. Perpendicularity: $\vec u\cdot\vec v=0$: $\ \lambda(2\lambda)+(-3)(-\lambda)+5(1)=0$.
   3. Simplify: $2\lambda^2+3\lambda+5=0$.
   4. Discriminant $=3^2-4(2)(5)=9-40=-31<0$.
   5. Since the discriminant is negative, there is no real value of $\lambda$ making the vectors perpendicular — the solution set is empty, $\varphi$.

   Common Mistakes

   • Forgetting to check the discriminant and assuming real roots always exist.
   • Sign slip while forming the dot product (e.g. writing $-3\lambda$ instead of $+3\lambda$).

   ✓Final answer

   The correct option is (D) — $\varphi$ (empty set).

   ANSWER: D

7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The plane passing through $(2,1,-3)$ and perpendicular to $3i-j+2k$ contains the points (A) $(1,5,1)$ & $(3,0,-5)$ (B) $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ & $\left(1,5,\dfrac{1}{2}\right)$ (C) $(3,1,-5)$ & $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ (D) $(1,5,3)$ & $(3,0,1)$

   ›Reveal solutionSolution

   This tests writing a plane's equation from a point and normal vector, then checking which pair of points satisfies it. The answer is option (B).

   Concept and Intuition

   A plane through point $P_0=(x_0,y_0,z_0)$ perpendicular to $\vec n=(a,b,c)$ has equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Once we have this equation, a point "lies in the plane" exactly when it satisfies the equation — we just substitute each candidate point and check.

   Step-by-Step Solution

   1. Normal vector $\vec n=3\hat i-\hat j+2\hat k=(3,-1,2)$, point $(2,1,-3)$.
   2. Plane equation: $3(x-2)-1(y-1)+2(z+3)=0 \Rightarrow 3x-6-y+1+2z+6=0 \Rightarrow 3x-y+2z+1=0$.
   3. Test option (A): $(1,5,1)$: $3(1)-5+2(1)+1=3-5+2+1=1\ne0$ — fails.
   4. Test option (B): $\left(\frac13,3,\frac12\right)$: $3\left(\frac13\right)-3+2\left(\frac12\right)+1=1-3+1+1=0$ — satisfies. $\left(1,5,\frac12\right)$: $3(1)-5+2\left(\frac12\right)+1=3-5+1+1=0$ — satisfies. Both points lie in the plane.
   5. Test option (C): $(3,1,-5)$: $9-1-10+1=-1\ne0$ — fails.
   6. Test option (D): $(1,5,3)$: $3-5+6+1=5\ne0$ — fails.
   7. So only option (B) has both its points satisfying the plane equation.

   Common Mistakes

   • Using the wrong sign convention when expanding the plane equation.
   • Checking only one of the two points in an option and stopping early.

   ✓Final answer

   The correct option is (B) — $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ & $\left(1,5,\dfrac{1}{2}\right)$.

   ANSWER: B

8. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.The number of vectors of unit length perpendicular to the two vectors $\vec{a} = (1,1,0)$ and $\vec{b} = (0,1,1)$ is (A) $1$ (B) $2$ (C) $3$ (D) Infinite

   ›Reveal solutionSolution

   The cross product gives one direction perpendicular to both vectors, and its two unit multiples ($+$ and $-$) are the only unit vectors satisfying the condition.

   Concept and Intuition

   Any vector perpendicular to two given non-parallel vectors must be a scalar multiple of their cross product; normalizing gives exactly two opposite unit vectors.

   Step-by-Step Solution

   1. $\vec a=(1,1,0)$, $\vec b=(0,1,1)$.
   2. $\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\\1&1&0\\0&1&1\end{vmatrix} = \hat i(1\cdot1-0\cdot1)-\hat j(1\cdot1-0\cdot0)+\hat k(1\cdot1-1\cdot0) = (1,-1,1)$.
   3. $|\vec a\times\vec b| = \sqrt{1+1+1}=\sqrt3$.
   4. Unit vectors perpendicular to both $\vec a$ and $\vec b$: $\pm\dfrac{1}{\sqrt3}(1,-1,1)$ — exactly two vectors.

   Common Mistakes

   • Forgetting the negative direction and answering "1" instead of "2".

   ✓Final answer

   The correct option is (B) — $2$.

   ANSWER: B

9. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.Let $\pi$ be the plane passing through the point (3, -3, 1) and perpendicular to the line joining the points (3, 4, -1) and (2, -1, 5). If the equation of the plane containing the points (3, 4, -1), (-1, 2, 5) and perpendicular to the plane $\pi$ is $ax + y + cz - d = 0$ then $3(a + c) =$ (A) $-d$ (B) $2d$ (C) $d$ (D) $-2d$

   ›Reveal solutionSolution

   Find $\pi$'s normal from the given perpendicular line, then build the required plane's normal as a cross product (perpendicular to $\pi$'s normal and lying along the given two points), and match coefficients.

   Concept and Intuition

   A plane through two points and perpendicular to another plane has a normal that must be perpendicular both to the direction joining the two points (since that line lies in the plane) and to the normal of the other plane (since the planes are perpendicular). That normal is exactly the cross product of those two vectors.

   Step-by-Step Solution

   1. Direction of the line joining $(3,4,-1)$ and $(2,-1,5)$: $(2-3,-1-4,5-(-1))=(-1,-5,6)$ — this is normal to $\pi$.
   2. $\pi$ through $(3,-3,1)$: $-1(x-3)-5(y+3)+6(z-1)=0\Rightarrow -x-5y+6z-18=0$, i.e. normal $\vec n_1=(1,5,-6)$ (up to sign).
   3. Direction joining $(3,4,-1)$ and $(-1,2,5)$: $(-4,-2,6)=\vec d$.
   4. Normal of the required plane: $\vec n_2=\vec n_1\times\vec d=(1,5,-6)\times(-4,-2,6)$. Computing: $i:(5\cdot6-(-6)(-2))=18$; $j:-(1\cdot6-(-6)(-4))=18$; $k:(1\cdot(-2)-5(-4))=18$. So $\vec n_2=(18,18,18)\parallel(1,1,1)$.
   5. Plane through $(3,4,-1)$ with normal $(1,1,1)$: $x+y+z=6$, i.e. $x+y+z-6=0$. Matching $ax+y+cz-d=0$: $a=1,c=1,d=6$.
   6. Verify it also contains $(-1,2,5)$: $-1+2+5-6=0$ ✓, and is $\perp\pi$: $(1)(1)+(1)(5)+(1)(-6)=0$ ✓.
   7. $3(a+c)=3(2)=6=d$.

   Common Mistakes

   • Sign errors in the cross product components.
   • Forgetting to check that the found plane is normalized to match the given form $ax+y+cz-d=0$ (coefficient of $y$ must be exactly $1$).

   ✓Final answer

   The correct option is (C) — $d$.

   ANSWER: C

10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.A unit vector that is perpendicular to the vector $2\bar{i} - \bar{j} + 2\bar{k}$ and coplanar with the vectors $\bar{i} + \bar{j} - \bar{k}$ and $2\bar{i} + 2\bar{j} - \bar{k}$ is (A) $\dfrac{\bar{i} + 2\bar{j} + \bar{k}}{\sqrt{6}}$ (B) $\dfrac{3\bar{i} + 2\bar{j} - 2\bar{k}}{\sqrt{17}}$ (C) $\dfrac{2\bar{i} + 2\bar{j} - \bar{k}}{3}$ (D) $\dfrac{3\bar{i} + 2\bar{j} + 2\bar{k}}{\sqrt{17}}$

    ›Reveal solutionSolution

    Write the general coplanar combination of the two given vectors as $a\,p+b\,q$, impose perpendicularity to the third vector to pin down $a=0$, then normalize the resulting direction.

    Concept and Intuition

    "Coplanar with $p$ and $q$" means the target vector is some linear combination $a\,p+b\,q$ (this spans exactly the plane through the origin containing both). Imposing perpendicularity to a third given vector is then just one linear equation in $a,b$ — it typically forces a ratio (or here, forces one coefficient to vanish entirely), collapsing the family to a single direction, which we then normalize to a unit vector.

    Step-by-Step Solution

    1. General coplanar vector: $v=a(\bar i+\bar j-\bar k)+b(2\bar i+2\bar j-\bar k)=(a+2b)\bar i+(a+2b)\bar j+(-a-b)\bar k$.
    2. Require $v\perp(2\bar i-\bar j+2\bar k)$: $2(a+2b)-1(a+2b)+2(-a-b)=0$.
    3. Simplify: $(a+2b)(2-1)+2(-a-b)=(a+2b)-2a-2b=-a$.
    4. So the condition reduces to $-a=0\Rightarrow a=0$.
    5. With $a=0$: $v=b(2\bar i+2\bar j-\bar k)$, i.e. $v$ is parallel to $2\bar i+2\bar j-\bar k$.
    6. Magnitude of $2\bar i+2\bar j-\bar k$ is $\sqrt{4+4+1}=3$.
    7. Unit vector: $v=\dfrac{2\bar i+2\bar j-\bar k}{3}$ (the $b<0$ choice just flips sign, giving the same line — the listed option is the standard positive orientation).

    Common Mistakes

    • Forgetting that "coplanar with two vectors" means a linear combination of exactly those two (through the origin), not some other plane.
    • Algebra slip while combining the $(a+2b)$ terms in the dot product.

    ✓Final answer

    The correct option is (C) — $\dfrac{2\bar{i} + 2\bar{j} - \bar{k}}{3}$.

    ANSWER: C

11. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.Let $\bar{a} = 3\bar{i} - \bar{j} - \bar{k}$, $\bar{b} = \bar{i} + \bar{j} - 2\bar{k}$ and $\bar{c} = 2\bar{i} + 2\bar{j} + \bar{k}$. Let $\bar{d}$ be a vector such that $|\bar{d}| = \sqrt{2}$ units. If the vector $\bar{d}$ is coplanar with $\bar{a}, \bar{b}$ and perpendicular to $\bar{c}$, then $\bar{d} =$ (A) $\pm\dfrac{1}{5}(3\bar{i} - 5\bar{j} + 4\bar{k})$ (B) $\pm\dfrac{1}{5}(-4\bar{i} + 5\bar{j} - 3\bar{k})$ (C) $\pm\dfrac{1}{5}(3\bar{i} + 5\bar{j} - 4\bar{k})$ (D) $\pm\dfrac{1}{5}(-3\bar{i} + 5\bar{j} + 4\bar{k})$

    ›Reveal solutionSolution

    $\bar d$ coplanar with $\bar a,\bar b$ means $\bar d=x\bar a+y\bar b$; perpendicularity to $\bar c$ fixes the ratio $x:y$; the given magnitude fixes the scale. The answer is (A).

    Concept and Intuition

    "Coplanar with $\bar a,\bar b$" means $\bar d$ lies in the plane spanned by $\bar a$ and $\bar b$, so it can be written as a linear combination $\bar d=x\bar a+y\bar b$ for some scalars $x,y$ (this is exactly what "coplanar with two given vectors, through the origin" means). The perpendicularity condition $\bar d\cdot\bar c=0$ then gives one constraint relating $x$ and $y$, so $\bar d$ is pinned down up to a single scalar multiple — which the given magnitude $|\bar d|=\sqrt2$ finally fixes (up to sign, since both directions along that line satisfy all the stated conditions).

    Step-by-Step Solution

    1. Given $\bar a=(3,-1,-1)$, $\bar b=(1,1,-2)$, $\bar c=(2,2,1)$.
    2. Since $\bar d$ is coplanar with $\bar a,\bar b$, write $\bar d = x\bar a + y\bar b = (3x+y,\,-x+y,\,-x-2y)$.
    3. Perpendicularity to $\bar c$: $\bar d\cdot\bar c=0$:

    $2(3x+y)+2(-x+y)+1(-x-2y)=0$

    $6x+2y-2x+2y-x-2y=0 \implies 3x+2y=0 \implies y=-\tfrac{3x}{2}.$

    4. Substitute back:

    $\bar d = \left(3x-\tfrac{3x}{2},\ -x-\tfrac{3x}{2},\ -x+3x\right) = \left(\tfrac{3x}{2},\,-\tfrac{5x}{2},\,2x\right).$

    Let $x=2t$ to clear fractions: $\bar d = (3t,-5t,4t) = t(3,-5,4)$.

    5. Apply $|\bar d|=\sqrt2$: $|\bar d|^2 = t^2(9+25+16)=50t^2=2 \implies t^2=\tfrac1{25}\implies t=\pm\tfrac15$.

    6. So $\bar d = \pm\dfrac15(3\bar i - 5\bar j + 4\bar k)$.

    Common Mistakes

    • Forgetting that "coplanar with $\bar a,\bar b$" through the origin means $\bar d$ is a linear combination of $\bar a,\bar b$ only — not involving $\bar c$ at all (that's reserved for the perpendicularity condition).
    • Dropping the $\pm$ sign — both directions satisfy every condition given, so both must appear in the final vector.

    ✓Final answer

    The correct option is (A) — $\pm\dfrac15(3\bar i-5\bar j+4\bar k)$.

    ANSWER: A

12. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If $\bar{a}=2\bar{i}+3\bar{j}, \bar{b}=3\bar{j}+4\bar{k}$ and $\bar{c}=5\bar{i}+4\bar{k}$ are three vectors, then a vector which is perpendicular to $\bar{a}$ and $\bar{b}\times\bar{c}$ is (A) $45\bar{i}-30\bar{j}+15\bar{k}$ (B) $3\bar{i}-2\bar{j}+\bar{k}$ (C) $-30\bar{i}+20\bar{j}+4\bar{k}$ (D) $-45\bar{i}+30\bar{j}+4\bar{k}$

    ›Reveal solutionSolution

    This tests the vector-triple-product idea: a vector perpendicular to both $\bar a$ and $\bar b\times\bar c$ is simply $\bar a\times(\bar b\times\bar c)$.

    Concept and Intuition

    The cross product of any two vectors is perpendicular to both of them. So if we want a single vector perpendicular to $\bar a$ AND to $\bar b\times\bar c$, the natural candidate is $\bar a\times(\bar b\times\bar c)$ — it is perpendicular to $\bar a$ by definition of cross product, and perpendicular to $\bar b\times\bar c$ for the same reason. No need to invoke the full triple-product expansion formula; we just compute it directly.

    Step-by-Step Solution

    1. Given $\bar a=2\bar i+3\bar j+0\bar k$, $\bar b=0\bar i+3\bar j+4\bar k$, $\bar c=5\bar i+0\bar j+4\bar k$.
    2. Compute $\bar b\times\bar c=\begin{vmatrix}\bar i&\bar j&\bar k\\0&3&4\\5&0&4\end{vmatrix}$ $=\bar i(3\cdot4-4\cdot0)-\bar j(0\cdot4-4\cdot5)+\bar k(0\cdot0-3\cdot5)=12\bar i+20\bar j-15\bar k$.
    3. Compute $\bar a\times(\bar b\times\bar c)=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&0\\12&20&-15\end{vmatrix}$ $=\bar i(3\cdot(-15)-0\cdot20)-\bar j(2\cdot(-15)-0\cdot12)+\bar k(2\cdot20-3\cdot12)$ $=\bar i(-45)-\bar j(-30)+\bar k(4)=-45\bar i+30\bar j+4\bar k$.
    4. This matches option (D).

    Common Mistakes

    • Computing $\bar c\times\bar b$ or $(\bar b\times\bar c)\times\bar a$ instead — cross product is anti-commutative, so sign/order matters.
    • Arithmetic slips in the $3\times3$ determinant expansion (missing the $-$ sign on the $\bar j$ cofactor).

    ✓Final answer

    The correct option is (D) — $-45\bar{i}+30\bar{j}+4\bar{k}$.

    ANSWER: D