Chhaya Mathematics Class 12 – SN Dey Vector Triple Product Solutions
If you’re studying Class 12 Mathematics under the WB Board and using the SN Dey book, this post is for you. Here you’ll find complete step-by-step solutions to all questions on the Vector Triple Product — explained clearly, with proper vector formulas and reasoning shown in each step.
The goal is simple: help you understand the logic behind every line, not just memorize answers. Each problem is solved in standard form using i, j, k notation and verified with the correct vector identities, so you can follow along easily while revising.
For the best experience, use desktop mode — it may take a few seconds to load because of the animations and detailed explanations built into each solution.
1(i). If ( vec{a}=hat{i}+hat{j}, vec{b}=hat{i}-hat{j}, vec{c}=5hat{i}+2hat{j}+3hat{k} ), find ( [vec{b} vec{c} vec{a}] ).
Write the scalar triple product as a determinant: [ [vec{b} vec{c} vec{a}] =begin{vmatrix} 1 & -1 & 0\[4pt] 5 & 2 & 3\[4pt] 1 & 1 & 0 end{vmatrix}. ]
Expand along the first row: [ =1cdot(2cdot0-3cdot1) -(-1)cdot(5cdot0-3cdot1) + 0cdot(cdots) ]
[ 1cdot(0-3) -(-1)cdot(0-3) = -3 – (+1)(-3) = -3 – 3 = -6. ]
Answer: ( [vec{b} vec{c} vec{a}] = -6. )
1(ii). If ( vec{alpha}=hat{i}-2hat{j}+3hat{k}, vec{beta}=2hat{i}-3hat{j}+hat{k}, vec{gamma}=3hat{i}+hat{j}-2hat{k} ), find ( vec{alpha}cdot(vec{beta}timesvec{gamma}) ).
Compute ( vec{beta}timesvec{gamma} ) using the determinant: [ vec{beta}timesvec{gamma} =begin{vmatrix} hat{i} & hat{j} & hat{k}\[4pt] 2 & -3 & 1\[4pt] 3 & 1 & -2 end{vmatrix}. ]
[ vec{beta}timesvec{gamma} = 5hat{i} + 7hat{j} + 11hat{k} ]
Now dot with ( vec{alpha} ): [ vec{alpha}cdot(vec{beta}timesvec{gamma}) =1cdot5 + (-2)cdot7 + 3cdot11 = 5 – 14 + 33 = 24. ]
Answer: ( vec{alpha}cdot(vec{beta}timesvec{gamma}) = 24. )
1(iii). If ( vec{a}=-hat{i}+2hat{j}+hat{k}, vec{b}=3hat{i}+hat{j}+2hat{k}, vec{c}=2hat{i}+hat{j}+3hat{k} ), find ( [vec{c} vec{a} vec{b}] ).
Determinant: [ [vec{c} vec{a} vec{b}] =begin{vmatrix} 2 & 1 & 3\[4pt] -1 & 2 & 1\[4pt] 3 & 1 & 2 end{vmatrix}. ]
Expand along the first row: [ =2(2cdot2 – 1cdot1) – 1((-1)cdot2 – 1cdot3) + 3((-1)cdot1 – 2cdot3). ]
[ 2(4-1) -1(-2-3) + 3(-1-6) = 2cdot3 -1(-5) + 3(-7) = 6 + 5 – 21 = -10. ]
Answer: ( [vec{c} vec{a} vec{b}] = -10. )
2(i). Vectors: ( hat{i}+hat{j}+hat{k}, hat{k}, 3hat{i}-hat{j}+2hat{k} ). Find the volume of the parallelepiped.
Volume = absolute scalar triple product: [ V = left| begin{vmatrix} 1 & 1 & 1\[4pt] 0 & 0 & 1\[4pt] 3 & -1 & 2 end{vmatrix} right|. ]
Expand (first row): [ =1(0cdot2 – 1cdot(-1)) – 1(0cdot2 – 1cdot3) + 1(cdots) ] [ =1(1) – 1(-3) = 1 + 3 = 4. ]
Volume = (4) cubic units.
2(ii). Vectors: ( 2hat{i}-hat{j}+hat{k}, hat{i}+2hat{j}-3hat{k}, 3hat{i}-4hat{j}+5hat{k} ). Find volume.
Determinant: [ begin{vmatrix} 2 & -1 & 1\[4pt] 1 & 2 & -3\[4pt] 3 & -4 & 5 end{vmatrix}. ]
Expand and compute term-by-term: [ =2(2cdot5 – (-3)(-4)) -(-1)(1cdot5 – (-3)cdot3) + 1(1cdot(-4)-2cdot3) ] [ =2(10-12) + (5+9) + (-4-6) = 2(-2) + 14 -10 = -4 + 14 -10 = 0. ]
Volume = (0) → vectors are coplanar.
2(iii). Vectors: ( hat{i}+hat{j}+hat{k}, hat{i}+2hat{j}+2hat{k}, hat{i}-2hat{j}+4hat{k} ). Find volume.
Determinant: [ begin{vmatrix} 1 & 1 & 1\[4pt] 1 & 2 & 2\[4pt] 1 & -2 & 4 end{vmatrix}. ]
Expand: [ =1(2cdot4 – 2cdot(-2)) – 1(1cdot4 – 2cdot1) + 1(1cdot(-2) – 2cdot1) ] [ =1(8+4) – (4-2) + (-2 – 2) = 12 – 2 – 4 = 6. ]
Volume = (6) cubic units.
3(i). Show vectors ( 4hat{i}+2hat{j}+hat{k}, 2hat{i}-hat{j}+3hat{k}, 8hat{i}+7hat{k} ) are coplanar.
Determinant: [ begin{vmatrix} 4 & 2 & 1\[4pt] 2 & -1 & 3\[4pt] 8 & 0 & 7 end{vmatrix}. ]
Expand and compute: [ =4((-1)cdot7 – 3cdot0) – 2(2cdot7 – 3cdot8) + 1(2cdot0 – (-1)cdot8) ] [ =4(-7) -2(14-24) + 8 = -28 -2(-10) + 8 = -28 + 20 + 8 = 0. ]
Triple product (=0) ⇒ coplanar.
3(ii). If ( vec{a}=hat{i}+hat{j}-6hat{k}, vec{b}=hat{i}+3hat{j}+4hat{k}, vec{c}=2hat{i}+5hat{j}+3hat{k} ), find the volume of the parallelepiped determined by these vectors.
Determinant: [ begin{vmatrix} 1 & 1 & -6\[4pt] 1 & 3 & 4\[4pt] 2 & 5 & 3 end{vmatrix}. ]
Expand along first row: [ =1(3cdot3 – 4cdot5) – 1(1cdot3 – 4cdot2) + (-6)(1cdot5 – 3cdot2). ]
Compute pieces: [ 1(9-20) – (3-8) -6(5-6) = 1(-11) -(-5) -6(-1) = -11 + 5 + 6 = 0. ]
Volume = (0) ⇒ vectors are coplanar.
3(iii). Using the same ( vec{a},vec{b},vec{c} ), let ( vec{u}=vec{a}-2vec{b}+vec{c}, vec{v}=2vec{a}+vec{b}-2vec{c}, vec{w}=3vec{a}-vec{b}-vec{c} ). Find the volume of the parallelepiped formed by ( vec{u},vec{v},vec{w} ).
Express new column matrix as product: [ [vec{u} vec{v} vec{w}] = [vec{a} vec{b} vec{c}]cdot P, ] where [ P = begin{pmatrix} 1 & 2 & 3 \ -2 & 1 & -1 \ 1 & -2 & -1 end{pmatrix} ] (columns are coefficients for u,v,w).
Determinant property: [ det[vec{u} vec{v} vec{w}] = det[vec{a} vec{b} vec{c}]cdotdet(P). ] From 3(ii) we have ( det[vec{a} vec{b} vec{c}] = 0 ). Therefore the product is (0).
Volume = (0). (Any linear combinations of coplanar vectors remain coplanar.)
4. If ( vec{a}=-2hat{i}-2hat{j}+4hat{k}, vec{b}=-2hat{i}+4hat{j}-2hat{k}, vec{c}=4hat{i}-2hat{j}-2hat{k} ), find ( vec{a}cdot(vec{b}timesvec{c}) ) and interpret.
Compute ( vec{b}timesvec{c} ): [ begin{vmatrix} hat{i} & hat{j} & hat{k}\[4pt] -2 & 4 & -2\[4pt] 4 & -2 & -2 end{vmatrix}. ]
So ( vec{b}timesvec{c} = -12hat{i} -12hat{j} -12hat{k}. )
Dot with ( vec{a} ): [ vec{a}cdot(vec{b}timesvec{c}) = (-2)(-12) + (-2)(-12) + 4(-12) = 24 + 24 – 48 = 0. ]
Answer: (0). Vectors are coplanar.
5. If ( xhat{i}-4hat{j}+5hat{k}, hat{i}+2hat{j}+hat{k}, 2hat{i}-hat{j}+hat{k} ) are coplanar, find ( x ).
Determinant zero: [ begin{vmatrix} x & -4 & 5\[4pt] 1 & 2 & 1\[4pt] 2 & -1 & 1 end{vmatrix} = 0. ]
Expand: [ = x(2cdot1 – 1cdot(-1)) – (-4)(1cdot1 – 1cdot2) + 5(1cdot(-1) – 2cdot2). ]
Compute: [ = x(2+1) + 4(1-2) + 5(-1-4) = 3x -4 -25 = 3x – 29. ] So (3x-29=0 Rightarrow x = dfrac{29}{3}.)
Answer: ( x = dfrac{29}{3} ).
6. If ( vec{a}=2hat{i}-hat{j}+hat{k}, vec{b}=hat{i}+2hat{j}-3hat{k}, vec{c}=3hat{i}+lambdahat{j}+5hat{k} ) are coplanar, find ( lambda ).
Determinant: [ begin{vmatrix} 2 & -1 & 1\[4pt] 1 & 2 & -3\[4pt] 3 & lambda & 5 end{vmatrix} = 0. ]
Expand: [ = 2(2cdot5 – (-3)lambda) -(-1)(1cdot5 – (-3)cdot3) + 1(1cdotlambda – 2cdot3). ]
Compute: [ =2(10+3lambda) + (5+9) + (lambda – 6) = 20 +6lambda +14 +lambda -6 = 28 + 7lambda. ] So (28 + 7lambda = 0 Rightarrow lambda = -4.)
Answer: ( lambda = -4 ).
7(i). Position vectors ( vec{A}=6hat{i}-4hat{j}+10hat{k}, vec{B}=-5hat{i}+3hat{j}-10hat{k}, vec{C}=4hat{i}-6hat{j}-10hat{k}, vec{D}=0hat{i}+2hat{j}+10hat{k} ). Prove coplanar.
Form ( overrightarrow{AB}=vec{B}-vec{A}=(-11,7,-20) ), ( overrightarrow{AC}=vec{C}-vec{A}=(-2,-2,-20) ), ( overrightarrow{AD}=vec{D}-vec{A}=(-6,6,0) ).
Compute triple product: [ begin{vmatrix} -11 & 7 & -20\[4pt] -2 & -2 & -20\[4pt] -6 & 6 & 0 end{vmatrix} ] Expand and evaluate → result (=0).
Triple product (=0) ⇒ A,B,C,D are coplanar.
7(ii). Position vectors ( vec{A}=4hat{i}+8hat{j}+12hat{k}, vec{B}=2hat{i}+4hat{j}+6hat{k}, vec{C}=3hat{i}+5hat{j}+4hat{k}, vec{D}=5hat{i}+8hat{j}+5hat{k} ). Prove coplanar.
Form AB, AC, AD and compute triple product (expansion shown earlier). Result (=0).
Hence A,B,C,D are coplanar.
8. If ( -4hat{i}-6hat{j}-2hat{k}, -hat{i}+4hat{j}+3hat{k}, -8hat{i}-hat{j}+lambdahat{k} ) are coplanar, find ( lambda ).
( displaystyle begin{vmatrix} -4 & -6 & -2\[4pt] -1 & 4 & 3\[4pt] -8 & -1 & lambda end{vmatrix} = 0. )
Expand: [ (-4)(4lambda+3) -(-6)(-lambda+24) + (-2)(33)=0. ]
Simplify: [ -16lambda-12 + 6(-lambda+24) -66 = -16lambda-12 -6lambda+144 -66 = -22lambda+66 = 0. ]
Solve: ( -22lambda+66=0 Rightarrow lambda=3. )
Answer: ( lambda=3 ).
9. If ( vec{a}=2hat{i}-lambdahat{j}+3hat{k}, vec{b}=3hat{i}+2hat{j}-muhat{k}, vec{c}=hat{i}+hat{j}+hat{k} ) are coplanar, find ( mu ) in terms of ( lambda ).
Determinant: [ begin{vmatrix} 2 & -lambda & 3\[4pt] 3 & 2 & -mu\[4pt] 1 & 1 & 1 end{vmatrix} = 0. ]
Expand (first row) and simplify: [ 2(2+mu) + lambda(3+mu) + 3 = 0 Rightarrow lambdamu + 2mu + 3lambda + 7 = 0. ]
Factor ( mu ): [ mu(lambda+2) = -3lambda – 7 Rightarrow mu = dfrac{-3lambda – 7}{lambda + 2} quad (lambda neq -2). ]
Answer: ( mu = dfrac{-3lambda – 7}{lambda + 2} ) (for ( lambdaneq -2 )).
10. Prove (i) ( (mathbf{a}+mathbf{b})cdotbig((mathbf{b}+mathbf{c})times(mathbf{c}+mathbf{a})big)=2mathbf{a}cdot(mathbf{b}timesmathbf{c}) ). (ii) ( mathbf{a}cdot(mathbf{b}times(mathbf{c}+mathbf{d}))=mathbf{a}cdot(mathbf{b}timesmathbf{c})+mathbf{a}cdot(mathbf{b}timesmathbf{d}) ).
(i) Expand: [ (mathbf{b}+mathbf{c})times(mathbf{c}+mathbf{a}) =mathbf{b}timesmathbf{c}+mathbf{b}timesmathbf{a}+mathbf{c}timesmathbf{a}. ]
Dot with ( mathbf{a}+mathbf{b} ): [ (mathbf{a}+mathbf{b})cdot(mathbf{b}timesmathbf{c}) +(mathbf{a}+mathbf{b})cdot(mathbf{b}timesmathbf{a}) +(mathbf{a}+mathbf{b})cdot(mathbf{c}timesmathbf{a}). ]
Drop zero terms and use cyclicity: [ mathbf{a}cdot(mathbf{b}timesmathbf{c}) + mathbf{b}cdot(mathbf{c}timesmathbf{a}) = mathbf{a}cdot(mathbf{b}timesmathbf{c}) + mathbf{a}cdot(mathbf{b}timesmathbf{c}) = 2mathbf{a}cdot(mathbf{b}timesmathbf{c}). ]
(ii) Expand: [ mathbf{b}times(mathbf{c}+mathbf{d})=mathbf{b}timesmathbf{c}+mathbf{b}timesmathbf{d}, ] then dot with ( mathbf{a} ): [ mathbf{a}cdot(mathbf{b}times(mathbf{c}+mathbf{d})) =mathbf{a}cdot(mathbf{b}timesmathbf{c})+mathbf{a}cdot(mathbf{b}timesmathbf{d}). ]
Both identities proved.
11. Given ( boldsymbol{alpha}=lambdahat{i}+hat{j}+3hat{k}, boldsymbol{beta}=-hat{i}+2hat{j}+hat{k}, boldsymbol{gamma}=3hat{i}+hat{j}+2hat{k} ) and ( [alpha beta gamma]=-10 ). Find ( lambda ).
Determinant: [ begin{vmatrix} lambda & 1 & 3\[4pt] -1 & 2 & 1\[4pt] 3 & 1 & 2 end{vmatrix} = -10. ]
Expand and simplify: [ =lambda(4-1) -1(-2-3) + 3(-1-6) = 3lambda +5 -21 = 3lambda -16. ] So (3lambda -16 = -10 Rightarrow 3lambda = 6 Rightarrow lambda = 2.)
Answer: ( lambda = 2 ).
12. If ( ahat{i}+ahat{j}+chat{k}, hat{i}+hat{k}, chat{i}+chat{j}+bhat{k} ) are coplanar, show ( c^2=ab ).
Determinant: [ begin{vmatrix} a & a & c\[4pt] 1 & 0 & 1\[4pt] c & c & b end{vmatrix}=0. ]
Expand: [ a(0cdot b – 1cdot c) – a(1cdot b – 1cdot c) + c(1cdot c – 0cdot c). ] Simplify: [ -ac – a(b-c) + c^2 = c^2 – ab. ]
Hence ( c^2 – ab = 0 Rightarrow c^2 = ab. )
13. Let ( mathbf{a}=hat{i}+hat{j}+hat{k}, mathbf{b}=hat{i}, mathbf{c}=c_1hat{i}+c_2hat{j}+c_3hat{k} ). If ( c_1=1, c_2=2 ), find ( c_3 ) so ( mathbf{a},mathbf{b},mathbf{c} ) are coplanar.
Determinant: [ begin{vmatrix} 1 & 1 & 1\[4pt] 1 & 0 & 0\[4pt] 1 & 2 & c_3 end{vmatrix}=0. ]
Expand: [ 1(0cdot c_3 – 0cdot 2) – 1(1cdot c_3 – 0cdot1) + 1(1cdot2 – 0cdot1) = -c_3 + 2. ] So ( -c_3 + 2 = 0 Rightarrow c_3 = 2. )
Answer: ( c_3 = 2 ).
14. Find ( x ) such that the four points ( A(3,2,1), B(4,x,5), C(4,2,-2), D(6,5,-1) ) are coplanar.
We find ( overrightarrow{AB}, overrightarrow{AC}, overrightarrow{AD} ) in vector form: [ overrightarrow{AB} = vec{B} – vec{A} = (4-3)hat{i} + (x-2)hat{j} + (5-1)hat{k} = hat{i} + (x-2)hat{j} + 4hat{k}. ] [ overrightarrow{AC} = vec{C} – vec{A} = (4-3)hat{i} + (2-2)hat{j} + (-2-1)hat{k} = hat{i} + 0hat{j} – 3hat{k}. ] [ overrightarrow{AD} = vec{D} – vec{A} = (6-3)hat{i} + (5-2)hat{j} + (-1-1)hat{k} = 3hat{i} + 3hat{j} – 2hat{k}. ]
For coplanarity: [ [overrightarrow{AB} overrightarrow{AC} overrightarrow{AD}] = 0. ] Substitute their components: [ begin{vmatrix} 1 & x-2 & 4\[4pt] 1 & 0 & -3\[4pt] 3 & 3 & -2 end{vmatrix} = 0. ]
Expand along the first row: [ = 1(0times(-2) – (-3)times3) – (x-2)(1times(-2) – (-3)times3) + 4(1times3 – 0times3). ]
[ = 1(9) – (x-2)(7) + 4(3). ]
Expand: [ 9 – 7(x-2) + 12 = 9 – 7x + 14 + 12 = 35 – 7x. ]
For coplanarity: [ 35 – 7x = 0 quadRightarrowquad x = 5. ]
Answer: ( boxed{x = 5} ).