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Hard Scalar and vector products MCQs for JEE

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Hard Scalar and vector products MCQs train multi-concept problem solving — attempt after mastering easy and medium sets (17+ on Goodmarks).

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Part of Vector Algebra in JEE Main Mathematics.

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Q1MathsUnit 12: Vector Algebra
Let a,β,γa, \beta, \gamma be distinct real numbers. The points with position vectors ai +βj++\beta j+ γk,βi+γj+ak,γi+aj+βk\gamma k, \beta i+\gamma j+a k, \gamma i+a j+\beta k
Q2MathsUnit 12: Vector Algebra
The non zero vector a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} related by a⃗=8b⃗\vec{a}=8 \vec{b} and c⃗=−7b⃗,\vec{c}=-7 \vec{b}, then angle between a⃗&c⃗\vec{a} \& \vec{c} is
Q3MathsUnit 12: Vector Algebra
Consider ΔABC\Delta A B C with A≡(a⃗),B≡(b⃗)A \equiv(\vec{a}), B \equiv(\vec{b}) and C=(c⃗),C=(\vec{c}), ff b⃗.(a⃗+c⃗)=b⃗⋅b⃗+\vec{b} .(\vec{a}+\vec{c})=\vec{b} \cdot \vec{b}+ a⃗⋅c⃗;∣b⃗−a⃗∣=3;∣c⃗−b⃗∣=4\vec{a} \cdot \vec{c} ;|\vec{b}-\vec{a}|=3 ;|\vec{c}-\vec{b}|=4 then the angle between the medians AM‾\overline{A M} and BDB D is
Q4MathsUnit 12: Vector Algebra
If position vectors of four vertices of a square are A=3i−λj,B=9i−6k\boldsymbol{A}=\mathbf{3} \boldsymbol{i}-\boldsymbol{\lambda} \boldsymbol{j}, \boldsymbol{B}=\mathbf{9} \boldsymbol{i}-\boldsymbol{6} \boldsymbol{k} C=6j−5k\boldsymbol{C}=\mathbf{6} \boldsymbol{j}-\mathbf{5} \boldsymbol{k} and D=αi−3j+4k\boldsymbol{D}=\boldsymbol{\alpha} \boldsymbol{i}-\boldsymbol{3} \boldsymbol{j}+\boldsymbol{4} \boldsymbol{k} then value of λ−α\lambda-\alpha is :
Q5MathsUnit 12: Vector Algebra
For any four points P,Q,R,SP, Q, R, S ∣PQ→×R→S−QR→×P→S+R→P×Q→S∣|\overrightarrow{P Q} \times \overrightarrow{\boldsymbol{R}} \boldsymbol{S}-\overrightarrow{\boldsymbol{Q} \boldsymbol{R}} \times \overrightarrow{\boldsymbol{P}} \boldsymbol{S}+\overrightarrow{\boldsymbol{R}} \boldsymbol{P} \times \overrightarrow{\boldsymbol{Q}} \boldsymbol{S}| is equal to 4 times the area of the triangle.
Q6MathsUnit 12: Vector Algebra
If A=(1,2,3),B=(2,3,4)\boldsymbol{A}=(\mathbf{1}, \mathbf{2}, \mathbf{3}), \boldsymbol{B}=(\mathbf{2}, \mathbf{3}, \mathbf{4}) and AB\boldsymbol{A} \boldsymbol{B} is produced upto CC such that 2AB=BC2 A B=B C then C=C=
Q7MathsUnit 12: Vector Algebra
If a,b,ca, b, c are non-coplanar vectors and λ\lambda is a real number, then [λ(a⃗+b⃗)λ2b⃗λc⃗]=[a⃗b⃗+c⃗b⃗]\left[\lambda(\vec{a}+\vec{b}) \lambda^{2} \vec{b} \lambda \vec{c}\right]=[\vec{a} \quad \vec{b}+\vec{c} \quad \vec{b}] for
Q8MathsUnit 12: Vector Algebra
The points A(−1,3,0),B(2,2,1)A(-1,3,0), B(2,2,1) and C(1,1,3)C(1,1,3) determine a plane. The distance of the plane A,B,CA, B, C from the point D(5,7,8)D(5,7,8) is

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