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Scalar and vector products MCQs for JEE

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Scalar and vector products JEE MCQs on Goodmarks include 7+ multiple-choice questions with correct answers and step-by-step solutions. Attempt free samples below or unlock the full bank with Pro.

Master Scalar and vector products through exam-style multiple-choice questions. This page features 7+ JEE MCQs covering Scalar and vector products, each with verified answers and clear explanations.

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Q1MathsUnit 12: Vector Algebra
Consider ΔABC\Delta A B C with A(a),B(b)A \equiv(\vec{a}), B \equiv(\vec{b}) and C=(c),C=(\vec{c}), ff b.(a+c)=bb+\vec{b} .(\vec{a}+\vec{c})=\vec{b} \cdot \vec{b}+ ac;ba=3;cb=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
Q2MathsUnit 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
Q3MathsUnit 12: Vector Algebra
The vectors i^+2j^+3k^,2i^j^+k^\hat{\boldsymbol{i}}+\boldsymbol{2} \hat{\boldsymbol{j}}+\boldsymbol{3} \hat{\boldsymbol{k}}, \boldsymbol{2} \hat{\boldsymbol{i}}-\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}} and 3i^+j^+4k^\mathbf{3} \hat{\mathbf{i}}+\hat{\boldsymbol{j}}+\mathbf{4} \hat{\boldsymbol{k}} are so placed that the end point of one vector is the starting point of the next vector. Then the vectors are :
Q4MathsUnit 12: Vector Algebra
In a triangle ABC,A B C, right angled at the vertex A,A, if the position vectors A,BA, B and CC are respectively 3i~+j~k~,i~+3 \tilde{i}+\tilde{j}-\tilde{k},-\tilde{i}+ 3j~+pk~\mathbf{3} \tilde{j}+p \tilde{k} and 5i~+qj~4k~,5 \tilde{i}+q \tilde{j}-4 \tilde{k}, then the point (p,q)(p, q) lies on a line
Q5MathsUnit 12: Vector Algebra
Given A1=2,A2=3\left|\overrightarrow{\boldsymbol{A}}_{1}\right|=2,\left|\overrightarrow{\boldsymbol{A}}_{2}\right|=\overrightarrow{\mathbf{3}} and A1+A2=3.\left|\vec{A}_{1}+\vec{A}_{2}\right|=3 . Find the value of (A1+2A2)(3A14A2)\left(\overrightarrow{\boldsymbol{A}}_{1}+\mathbf{2} \overrightarrow{\boldsymbol{A}}_{2}\right) \cdot\left(\boldsymbol{3} \overrightarrow{\boldsymbol{A}}_{1}-\boldsymbol{4} \boldsymbol{\vec { A }}_{2}\right)
Q6MathsUnit 12: Vector Algebra
Assertion Let aˉ,bˉ,rˉ\bar{a}, \bar{b}, \bar{r} be the vectors such that rˉ+\bar{r}+ r×a=bthenr2=(ab)2+b21+a2\overline{\boldsymbol{r}} \times \overline{\boldsymbol{a}}=\overline{\boldsymbol{b}} \operatorname{then}|\overline{\boldsymbol{r}}|^{2}=\frac{(\overline{\boldsymbol{a}} \cdot \overline{\boldsymbol{b}})^{2}+|\overline{\boldsymbol{b}}|^{2}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}} Reason r=(ab)a+b+a×b1+a2\overline{\boldsymbol{r}}=\frac{(\overline{\boldsymbol{a}} \cdot \overline{\boldsymbol{b}}) \overline{\boldsymbol{a}}+\overline{\boldsymbol{b}}+\overline{\boldsymbol{a}} \times \overline{\boldsymbol{b}}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}}
Q7MathsUnit 12: Vector Algebra
Equation of the plane containing the linesr=(i2j+k)+t(i+2jk)\operatorname{lines} \overline{\boldsymbol{r}}=(\overline{\boldsymbol{i}}-\boldsymbol{2} \overline{\boldsymbol{j}}+\overline{\boldsymbol{k}})+\boldsymbol{t}(\overline{\boldsymbol{i}}+\mathbf{2} \overline{\boldsymbol{j}}-\overline{\boldsymbol{k}}) r=(i+2jk)+s(i+j+3k)\boldsymbol{\boldsymbol { r }}=(\overline{\boldsymbol{i}}+\mathbf{2} \overline{\boldsymbol{j}}-\overline{\boldsymbol{k}})+\boldsymbol{s}(\overline{\boldsymbol{i}}+\overline{\boldsymbol{j}}+\mathbf{3} \overline{\boldsymbol{k}}) is

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